Questions And Answers In School Physics

Lev Tarasov Aldina Tarasova

QUESTIONS
AND ANSWERS
IN SCHOOL PHYSICS

MIR PUBLISHERS MOSCOW

LEV TARASOV, ALDINA TARASOVA

QUESTIONS

AND ANSWERS

IN SCHOOL PHYSICS

MIR PUBLISHERS MOSCOW

Translated from the Russian by Nicholas Weinstein.

First published 1973.

Revised from the 1968 Russian edition.

This electronic version typeset in RTpX by Damitr Mazanav.
Released on the web by http: //mirtitles .org in 2020.

Git Repository for obtaining the source files
https://gitlab.com/mirtitles/tarasov- gasp/

Contents

From The Editor Of The Russian Edition

Foreword 9

§ 1 Can You Analyse Graphs Representing The

Kinematics Of Straight-Line Motion? 13

§ 2 Can You Show The Forces Applied To A Body? 21

§ 3 Can You Determine The Friction Force? 31

§ 4 How Well Do You Know Newton's Laws Of Motion? 37

§ 5 How Do You Go About Solving Problems In

Kinematics? 51

Problems 59

§ 6 How Do You Go About Solving Problems In

Dynamics? 61

Problems 65

§ 7 Are Problems In Dynamics Much More Difficult

To Solve If Friction Is Taken Into Account? 67

Problems 73

§ 8 How Do You Deal With Motion In A Circle?

Problems 89

4

§ 9 How Do You Explain The Weightlessness Of Bodies? 91
Problems 95

§ 10 Can You Apply The Laws Of Conservation Of

Energy And Linear Momentum? 99

Problems 115

§77 Can You Deal With Harmonic Vibrations? 119

Problems 126

§72 What Happens To A Pendulum In A State Of

Weightlessness? 12 7

§ 13 Can You Use The Force Resolution Method

Efficiently? 135

Problems 138

§ 14 What Do You Know About The Equilibrium Of

Bodies? 141

§ 15 How Do You Locate The Centre Of Gravity? 149

Problems 155

§ 16 Do you know Archimedes’ principle? 159

Problems 165

§ 17 Is Archimedes’Principle Valid In A Spaceship? 167

% 18 What Do You Know About The Molecular-Kinetic

Theory Of Matter? 173

§ 19 How Do You Account For The Peculiarity In The

Thermal Expansion Of Water? 187

§ 20 How Well Do You Know The Gas Laws? 189

§27 How Do You Go About Solving Problems On

Gas Laws? 203

Problems 211

§ 22 Let Us Discuss Field Theory 215

5

§ 23 How Is An Electrostatic Field Described? 221

§ 24 How Do Lines Of Force Behave Near The Surface

Of A Conductor? 231

§ 25 How Do You Deal With Motion In A Uniform

Electrostatic Field? 237

Problems 247

§ 26 Can You Apply Coulomb’s Law? 249

Problems 256

§ 27 Do You Know Ohm’s Law? 259

Problems 267

§ 28 Can A Capacitor Be Connected Into A

Direct-Current Circuit? 269

Problems 271

§ 29 Can you compute the resistance of a branched

portion of a circuit? 275

Problems 280

§ 30 Why Did The Electric Bulb Burn Out? 283

Problems 288

% 31 Do You Know How Light Beams Are Reflected

And Refracted? 293

Problems 299

§ 32 How Do You Construct Images Formed By

Mirrors And Lenses? 301

§ 33 How Well Do You Solve Problems Involving

Mirrors A nd Lenses? 313

Problems 318

Answers 321

From The Editor Of
The Russian Edition

It can safely be asserted that no student preparing for an entrance ex¬
amination in physics, for admission to an engineering institute has yet
opened a book similar to this one. Employing the extremely lively
form of dialogue, the authors were able to comprehensively discuss
almost all the subjects in the syllabus, especially questions usually con¬
sidered difficult to understand. The book presents a detailed analysis
of common mistakes made by students taking entrance examinations
in physics. Students will find this to be an exceptionally clear and in¬
teresting textbook which treats of complicated problems from various
viewpoints and contains a great many excellent illustrations promoting
a deeper understanding of the ideas and concepts involved. The au¬
thors are lecturers of the Moscow Institute of Electronics Engineering
and are well acquainted with the general level of training of students
seeking admission to engineering institutes; they have years of experi¬
ence in conducting entrance examinations. The expert knowledge of
the authors, in conjunction with the lively and lucid presentation, has
made. This a very useful study guide for students preparing for physics
examinations.

Prof. G. Epifanov, D.Sc. (Phys. and Math.)

Foreword

This book was planned as an aid to students preparing for an entrance
examination in physics for admission to an engineering institute. It has
the form of a dialogue between the author (the TEACHER:) and an in¬
quisitive reader (the STUDENT:). This is exceptionally convenient for
analysing common errors made by students in entrance examinations,
for reviewing different methods of solving the same problems and for
discussing difficult questions of physical theory.

A great many questions and problems of school physics are dealt
with. Besides, problems are given (with answers) for home study. Most
of the questions and problems figured in the entrance examinations
of the Moscow Institute of Electronics Engineering in the years 1964-
66. An analysis of mistakes made by students is always instructive.
Attention can be drawn to various aspects of the problem, certain
fine points can be made, and a more thorough understanding of the
fundamentals can be reached.

Such an analysis, however, may prove to be very difficult. Though
there is only one correct answer, there can be a great many incorrect
ones. It is practically impossible to foresee all the incorrect answers to
any question; many of them remain concealed forever behind the dis¬
tressing silence of a student being orally examined. Nevertheless, one
can point out certain incorrect answers to definite questions that are
heard continually. There are many questions that are almost inevitably
answered incorrectly. This book is based mainly on these types of
questions and problems.

We wish to warn the reader that this is by no means a textbook em-

10

bracing all the items of the syllabus. He will not find here a systematic
account of the subject matter that may be required by the study course
in physics. He will find this text to be perhaps more like a freely told
story or, rather, a freely conducted discussion. Hence, it will be of
little use to those who wish to begin their study of physics or to sys¬
tematize their knowledge of this science. It was intended, instead, for
those who wish to increase their knowledge of physics on the thresh¬
old of their examinations.

Our ideal reader, as we conceive him, has completed the required
course in school physics, has a good general idea of what it is all about,
remembers the principal relationships, can cite various laws and has a
fair knowledge of the units employed. He is in that “suspended” state
in which he is no longer a secondary school student and has not yet
become a full fledged student of an institute. He is eager, however, to
become one. If this requires an extension of his knowledge in physics,
our book can help him.

Primarily, we hope our book will prove that memorizing a text¬
book (even a very good one) is not only a wearisome business, but
indeed a fruitless one. A student must learn to think, to ponder over
the material and not simply learn it by heart. If such an understand¬
ing is achieved, to some extent or other, we shall consider our efforts
worthwhile.

In conclusion, we wish to thank Prof. G. Epifanov without whose
encouragement and invaluable aid this book could not have been
written and prepared for publication. We also gratefully acknowledge
the many helpful suggestions and constructive criticism that were
made on the manuscript by Prof. V. A. Fabrikant, Associate-Prof.

A. G. Chertov, and E. N. Vtorov, Senior Instructor of the Physics
Department of the Moscow Power Engineering Institute.

L. Tarasov
A. Tarasova

Do not neglect kinematics! The question of how a body travels in space
and time is of considerable interest, both from physical and practical
points of view.

§ 1 Can You Analyse Graphs Representing
The Kinematics Of Straight-Line Motion?

TEACHER: You have seen graphs showing the dependence of the
velocity and distance travelled by a body on the time of travel for
straight-line, uniformly variable motion. In this connection, I
wish to put the following question. Consider a velocity graph of
the kind shown in Figure 1. On its basis, draw a graph showing
the dependence of the distance travelled on time.

STUDENT: But I have never drawn such graphs.

TEACHER: There should be no difficulties. However, let us rea¬
son this out together. First we will divide the whole interval of
time into three periods: 1, 2 and 3 (see Figure 1). How does the
body travel in period 1? What is the formula for the distance
travelled in this period?

Figure 1: On the basis of this
graph can you draw a graph
showing the dependence of
distance travelled on time?

STUDENT: In period 1, the body has uniformly accelerated mo¬
tion with no initial velocity. The formula for the distance trav-

14 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

elled is of the form

= ( 1 )

where a is the acceleration of the body.

TEACHER: Using the velocity graph, can you find the accelera¬
tion?

STUDENT: Yes. The acceleration is the change in velocity in unit
time. It equals the ratio of length AC to length OC.

TEACHER: Good. Now consider periods 2 and 3.

STUDENT: In period 2 the body travels with uniform velocity
v acquired at the end of period 1. The formula for the distance
travelled is s = vt.

TEACHER: Just a minute, your answer is inaccurate. You have
forgotten that the uniform motion began, not at the initial in¬
stant of time, but at the instant tj. Up to that time, the body had

at:

already travelled a distance equal to The dependence of the
distance travelled on the elapsed time for period 2 is expressed by
the equation

att

s(t) = -+v{t-t x )

( 2 )

With this in mind, please write the formula for the distance
travelled in period 3.

STUDENT: The motion of the body in period 3 is uniformly de¬
celerated. If I understand it correctly, the formula of the distance
travelled in this period should be

L a

s(t) = —+v(t 2 -t 1 ) + v(t-t 2 )

ll(t~t 2 ) 2

where a 1 is the acceleration in period 3. It is only one half of the
acceleration a in period 1, because period 3 is twice as long as
period 1.

CAN YOU ANALYSE GRAPHS REPRESENTING THE KINEMATICS OF STRAIGHT-LINE MOTION? 15

TEACHER: Your equation can be simplified somewhat:

= + --—

( 3 )

Now, it remains to summarize the results of equations (1), (2) and

( 3 ).

STUDENT: I understand. The graph of the distance travelled has
the form of a parabola for period 1, a straight line for period 2
and another parabola (but turned over, with the convexity facing
upward) for period 3. Here is the graph I have drawn (Figure 2).

Figure 2: Students’ incorrect
graph showing distance covered
s as a function of time t.

TEACHER: There are two faults in your drawing: the graph of the
distance travelled should have no kinks. It should be a smooth
curve, i.e. the parabolas should be tangent to the straight line.
Moreover, the vertex of the upper (inverted) parabola should
correspond to the instant of time f 3 . Here is a correct drawing of
the graph (Figure 3).

STUDENT: Please explain it.

TEACHER: Let us consider a portion of a distance-travelled vs
time graph (Figure 4). The average velocity of the body in the
interval from t to t + At equals

s(t + At) — s(t)

-= tan a

At

where a is the angle between chord AB and the horizontal. To
determine the velocity of the body at the instant t it is necessary

16 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 3: Corrected graph
showing distance covered s as a
function of time f.

to find the limit of such average velocities for At —> 0. Thus

v{t)

s{t + At) — s{t)

iim -

At—*0 At

( 4 )

In the limit, the chord becomes a tangent to the distance-travelled
vs time curve, passing through point A (see the dashed line in
Figure 4). The tangent of the angle this line (tangent to the curve)
makes with the horizontal is the value of the velocity at the
instant t . Thus it is possible to find the velocity at any instant of
time from the angle of inclination of the tangent to the distance-
travelled vs time curve at the corresponding point. But let us

Figure 4: Corrected graph
showing distance covered s as a
function of time t.

return to your drawing (see Figure 2). It follows from your
graph that at the instant of time fj (and at t 2 ) the velocity of
the body has two different values. If we approach t , from the

CAN YOU ANALYSE GRAPHS REPRESENTING THE KINEMATICS OF STRAIGHT-LINE MOTION? 17

left, the velocity equals tan a 1 , while if we approach it from the
right the velocity equals tan a 2 ■ According to your graph, the
velocity of the body at the instant f 3 (and again at t 2 ) must have
a discontinuity, which actually it has not (the velocity vs time
graph in Figure 1 is continuous).

STUDENT: I understand now. Continuity of the velocity graph
leads to smoothness of the distance-travelled vs time graph.

TEACHER: Incidentally, the vertices of the parabolas should cor¬
respond to the instants of time 0 and t 3 because at these instants
the velocity of the body equals zero and the tangent to the curve
must be horizontal for these points. Now, using the velocity
graph in Figure 1, find the distance travelled by a body by the
instant t 2 .

STUDENT: First we determine the acceleration a in period 1 from
the velocity graph and then the velocity v in period 2. Next
we make use of formula (2). The distance travelled by the body
during the time t 2 equals

S(t 2 ) = -y+v(t 2 ~t)

TEACHER: Exactly. But there is a simpler way. The distance
travelled by the body during the time t 2 is numerically equal to
the area of the figure OABD under the velocity vs time graph in
the interval Of ? . Let us consider another problem to fix what we
have learned.

Assume that the distance-travelled vs time graph has kinks. This
graph is shown in Figure 5, where the curved line is a parabola
with its vertex at point A. Draw the velocity vs timegraph.

STUDENT: Since there are kinks in the distance-travelled graph,
there should be discontinuities in the velocity graph at the cor¬
responding instants of time (tj and t 2 ). Flere is my graph (Fig¬
ure 6).

TEACHER: Good. What is the length of SC?

18 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 5: Corrected graph
with kinks showing distance
travelled s as a function of time
t.

Figure 6: Velocity vs time
graph for the function shown
in Figure 5.

STUDENT: It is equal to tan a 1 (see Figure 5). We don’t, however,
know the value of angle a 1 .

TEACHER: Nevertheless, we should have no difficulty in deter¬
mining the length of SC. Take notice that the distance travelled
by the body by the time f 3 is the same as if it had travelled at uni¬
form velocity all the time (the straight line in the interval from
t 2 to t 3 in Figure 5 is a continuation of the straight line in the
interval from 0 to fj). Since the distance travelled is measured
by the area under the velocity graph, it follows that the area of
rectangle AD EC in Figure 6 is equal to the area of triangle ABC.
Consequently, BC = 2EC, i.e. the velocity at instant f 9 when
approached from the left is twice the velocity of uniform motion
in the intervals from 0 to fj and from t 2 to f 3 .

The concept of a force is one of the basic physical concepts. Can you
apply it with sufficient facility? Do you have a good understanding of
the laws of dynamics?

m

§ 2 Can You Show The Forces Applied To A Body?

STUDENT: Problems in mechanics seem to be the most difficult
of all. How do you begin to solve them?

TEACHER: Frequently, you can begin by considering the forces
applied to a body. As an example, we can take the following cases
Figure 7:

(a) the body is thrown upward at an angle to the horizontal,

(b) the body slides down an inclined plane,

(c) the body rotates on the end of a string in a vertical plane,
and

(d) the body is a pendulum.

Draw arrows showing the forces applied to the body in each of
these cases, and explain what the arrows represent.

STUDENT: Here is my drawing (Figure 8). In the first case, P
is the weight of the body and F is the throwing force. In the
second, P is the weight, F is the force which keeps the body
sliding along the, plane and iy r is the friction force. In the third,
P is the weight, F c , is the centripetal force and T is the tension in
the string. In the fourth case, P is the weight, F is the restoring
force and T is the tension in the string.

TEACHER: You have made mistakes in all four cases. Here I have
the correct drawing (Figure 9). One thing that you must under¬
stand clearly is that a force is the result of interaction between

/ \
(a)

(b)

(d)

Figure 7: A variety of different
situations for bodies. For each
case we have to find out the
forces acting on the given body.

22 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

bodies. Therefore, to show the forces applied to a body you must
first establish what bodies interact with the given body. Thus, in
the first case, only the earth interacts with the body by attract¬
ing it (Figure 9 (a)). Therefore, only one force, the weight P, is
applied to the body. If we wished to take into consideration the
resistance of the air or, say, the action of the wind, we would
have to introduce additional forces. No “throwing force”, shown
in your drawing, actually exists, since there is no interaction
creating such a force.

STUDENT: But to throw a body, surely some kind of force must
be exerted on it.

TEACHER: Yes, that’s true. When you throw a body you exert
a certain force on it. In the case above, however, we dealt with
the motion of the body after it was thrown, i. e. after the force
which imparted a definite initial velocity of flight to the body
had ceased to act. It is impossible to “accumulate” forces; as soon
as the interaction of the bodies ends, the force isn’t there any
more.

STUDENT: But if only the weight is acting on the body, why
doesn’t it fall vertically downward instead of travelling along a
curved path?

TEACHER: It surprises you that in the given case the direction
of motion of the body does not coincide with the direction of
the force acting on it. This, however, fully agrees with Newton’s
second law. Your question shows that you haven’t given suffi¬
cient thought to Newton’s laws of dynamics. I intend to discuss
this later (see § 4). Now I want to continue our analysis of the
four cases of motion of a body. In the second case (Figure 9 (b)), a
body is sliding down an inclined plane. What bodies are interact¬
ing with it?

STUDENT: Evidently, two bodies: the earth and the inclined
plane.

TEACHER: Exactly. This enables us to find the forces applied
to the body. The earth is responsible for the weight P, and the

7^77777777777777777777777/7777^77

(a)

(d)

Figure 8: Student response
for forces acting on bodies in
different situations given in
Figure 7.

CAN YOU SHOW THE FORCES APPLIED TO A BODY? 23

inclined plane causes the force of sliding friction F fr and the

force N ordinarily called the bearing reaction 1 . Note that you 1 Also known as the normal

entirely omitted force N in your drawing. force.

STUDENT: Just a moment! Then the inclined plane acts on the

A

ren /

V - '-X

s

\

body with two forces and not one?

777777777777777777/72

P

TEACHER: There is, of course, only one force. It is, however,
more convenient to deal with it in the form of two component
forces, one directed along the inclined plane (force of sliding
friction) and the other perpendicular to it (bearing reaction). The
fact that these forces have a common origin, i.e, that they are
components of the same force, can be seen in the existence of a
universal relation between F fr and N:

Ffr = kN (5)

where k is a constant called the coefficient of sliding friction. We
shall deal with this relationship in more detail later (§ 3).

STUDENT: In my drawing, I showed a sliding force which keeps
the body sliding down the plane. Evidently, there is no such
force. But I clearly remember hearing t.he term “sliding force”
used frequently in the past. What can you say about this?

Figure 9: For each case the
correct drawing of forces
acting on the bodies are made.
Compare this with Figure 8.

TEACHER: Yes, such a term actually exists. You must bear in
mind, however, that the sliding force, as you call it, is simply
one of the components of the body’s weight, obtained when
the weight is resolved into two forces, one along the plane and
the other normal to it If, in enumerating the forces applied
to the body, you have named the weight, there is no reason to
add the sliding force, one of its components. In the third case
(Figure 9 (c)), the body rotates in a vertical plane. What bodies
act on it?

STUDENT: Two bodies: the earth and the string.

TEACHER: Good, and that is why two forces are applied to the
body: the weight and the tension of the string.

STUDENT: But what about the centripetal force?

24 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: Don’t be in such a hurry! So many mistakes are
made in problems concerning the motion of a body in a circle
that I intend to dwell at length on this further on (§ 8). Here
I only wish to note that the centripetal force is not some kind
of additional force applied to the body. It is the resultant force.

In our case (when the body is at the lowest point of its path),
the centripetal force is the difference between the tension of the
string and the weight.

STUDENT: If I understand it correctly, the restoring force in the
fourth case (Figure 9 (d)) is also the resultant of the tension in the
string and the weight?

TEACHER: Quite true. Here, as in the third case, the string and
the earth interact with the body. Therefore, two forces, the
tension of the string and the weight, are applied to the body.

I wish to emphasize again that forces arise only as a result of
interaction of bodies; they cannot originate from any “accessory”
considerations. Find the bodies acting on the given object and
you will reveal the forces applied to the object.

STUDENT: No doubt there are more complicated cases than the
ones you have illustrated in (Figure 7). Can we consider them?

TEACHER: There are many examples of more complicated inter¬
action of bodies. For instance, a certain constant horizontal force
F acts on a body as a result of which the body moves upward
along an inclined surface. The forces applied to the body in this
case are shown in (Figure 10).

N

Figure 10: A body on inclined
plane and forces acting on it.

CAN YOU SHOW THE FORCES APPLIED TO A BODY? 25

Another example is the oscillation of an electrically charged pen¬
dulum placed inside a parallel-plate capacitor. Here we have an
additional force F e with which the field of the capacitor acts on
the charge of the pendulum (Figure 11). It is obviously impos¬
sible to mention all the conceivable cases that may come up in
solving problems.

Figure 11: Forces acting on an
electrically charged pendulum
inside a parallel plate capacitor.

STUDENT: What do you do when there are several bodies in the
problem? Take, for example, the case illustrated in Figure 12.

TEACHER: You should clearly realize each time the motion of
what bodies or combination of bodies you intend to consider.
Let us take, for instance, the motion of body 1 in the example
you proposed. The earth, the inclined plane and string AB inter¬
act with this body.

Figure 12: What are the forces
acting on three bodies on
balanced on an incline?

STUDENT: Doesn’t body 2 interact with body 1?

TEACHER: Only through string AS. The forces applied to body
1 are the weight P ', force Fjr r of sliding friction, bearing reaction
N' and the tension T' of string AS (Figure 13 (a)).

26 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT: But why is the friction force directed to the left in
your drawing? It would seem just as reasonable to have it act in
the opposite direction.

Figure 13: Forces acting on the
three bodies balanced on an
incline as shown in Figure 12.

TEACFIER: To determine the direction of the friction force, it is
necessary to know the direction in which the body is travelling.

If this has not been specified in the problem, we should assume
either one or the other direction. In the given problem, I assume
that body 1 (together with the whole system of bodies) is travel¬
ling to the right and the pulley is rotating clockwise. Of course,

I cannot know this beforehand; the direction of motion becomes
definite only after the corresponding numerical values are sub¬
stituted. If my assumption is wrong, I shall obtain a negative
value when I calculate the acceleration. Then I will have to as¬
sume that the body moves to the left instead of to the right (with
the pulley rotating counterclockwise) and to direct the force of
sliding friction correspondingly. After this I can derive an equa¬
tion for calculating the acceleration and again check its sign by
substituting the numerical values.

CAN YOU SHOW THE FORCES APPLIED TO A BODY? 27

STUDENT: Why check the sign of the acceleration a second time?
If it was negative when motion was assumed to be to the right, it
will evidently be positive for the second assumption.

TEACHER: No, it can turn out to be negative in the second case as
well.

STUDENT: I can’t understand that. Isn’t it obvious that if the
body is not moving to the right it must be moving to the left?

TEACHER: You forget that the body can also be at rest. We shall
return to this question later and analyse in detail the complica¬
tions that arise when we take the friction force into considera¬
tion (see § 7).

For the present, we shall just assume that the pulley rotates
clockwise and examine the motion of body 2.

STUDENT: The earth, the inclined plane, string AB and string
CD interact with body 2. The forces applied to body 2 are
shown in Figure 13 (b).

TEACHER: Very well. Now let us go over to body 3.

STUDENT: Body 3 interacts only with the earth and with string
CD. Figure 13 (c) shows the forces applied to body 3.

TEACHER: Now, after establishing the forces applied to each
body, you can write the equation of motion for each one and
then solve the system of equations you obtain.

STUDENT: You mentioned that it was not necessary to deal with
each body separately, but that we could also consider the set of
bodies as a whole.

TEACHER: Why yes; bodies 1, 2 and 3 can be examined, not sep¬
arately as we have just done, but as a whole. Then, the tensions
in the strings need not be taken into consideration since they
become, in this case, internal forces, i.e. forces of interaction be¬
tween separate parts of the item being considered. The system of
the three bodies as a whole interacts only with the earth and the
inclined plane.

28 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT: I should like to clear up one point. When I depicted
the forces in Figure 13 (b) and (c), I assumed that the tension in
string CD is the same on both sides of the pulley. Would that be
correct?

TEACHER: Strictly speaking, that’s incorrect. If the pulley is
rotating clockwise, the tension in the part of string CD attached
to body 3 should be greater than the tension in the part of the
string attached to body 2. This difference in tension is what
causes accelerated rotation of the pulley. It was assumed in the
given example that the mass of the pulley can be disregarded. In
other words, the pulley has no mass that is to be accelerated, it is
simply regarded as a means of changing the direction of the string
connecting bodies 2 and 3. Therefore, it can be assumed that
the tension in string CD is the same on both sides of the pulley.
As a rule, the mass of the pulley is disregarded unless otherwise
stipulated. Have we cleared up everything?

STUDENT: I still have a question concerning the point of applica¬
tion of the force. In your drawings you applied all the forces to a
single point of the body. Is this correct? Can you apply the force
of friction, say, to the centre of gravity of the body?

TEACHER: It should be remembered that we are studying the
kinematics and dynamics, not of extended bodies, but of mate¬
rial points, or particles, i.e. we regard the body to be of point
mass. On the drawings, however, we show a body, and not a
point, only for the sake of clarity. Therefore, all the forces can be
shown as applied to a single point of the body.

STUDENT: We were taught that any simplification leads to the
loss of certain aspects of the problem. Exactly what do we lose
when we regard the body as a material point?

TEACHER: In such a simplified approach we do not take into ac¬
count the rotational moments which, under real conditions, may
result in rotation and overturning of the body. A material point
has only a motion of translation. Let us consider an example.
Assume that two forces are applied at two different points of a
body: F { at point A and D> at point B, as shown in Figure 14 (a).

CAN YOU SHOW THE FORCES APPLIED TO A BODY? 29

Figure 14: Forces acting on the
three bodies on balanced on an
incline shown in Figure 12.

Now let us apply, at point A, force F 2 equal and parallel to force
F 2 , and also force F" equal to force F 2 but acting in the opposite
direction (see Figure 14 (b)). Since forces F 2 and F 2 counterbal¬
ance each other, their addition does not alter the physical aspect
of the problem in any way. However, Figure 14 (b) can be in¬
terpreted as follows: forces F x and F 2 applied at point A cause
motion of translation of the body also applied to the body is a
force couple (forces F 2 and F") causing rotation. In other words,
force F 2 can be transferred to point A of the body if, at the same
time, the corresponding rotational moment is added. When
we regard the body as a material point, or particle, there will
evidently be no rotational moment.

STUDENT: You say that a material point cannot rotate but has
only motion of translation. But we have already dealt with
rotational motion - motion in a circle.

TEACHER: Do not confuse entirely different things. The motion
of translation of a point can take place along various paths, for
instance in a circle. When I ruled out the possibility of rotational
motion of a point I meant rotation about itself, i.e, about any
axis passing through the point.

§ 3 Can You Determine The Friction Force?

TEACHER: I should like to dwell in more detail on the calculation
of the friction force in various problems. I have in mind dry
sliding friction (sliding friction is said to be dry when there is no
layer of any substance, such as a lubricant, between the sliding
surfaces).

STUDENT: But here everything seems to be quite clear.

Figure 15: A sled being pulled
with a rope.

TEACHER: Nevertheless, many mistakes made in examinations
are due to the inability to calculate the friction force. Consider
the example illustrated in Figure 15. A sled of weight P is being
pulled with a force F applied to a rope which makes an angle
a with the horizontal; the coefficient of friction is k. Find the
force of sliding friction. How will you go about it?

STUDENT: Why, that seems to be very simple. The friction force
equals kP.

32 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: Entirely wrong. The force of sliding friction is equal,
not to kP, but to kN, where N is the bearing reaction. Remem¬
ber equation (5) from § 2.

STUDENT: But isn’t that the same thing?

TEACHER: In a particular case, the weight and the bearing reac¬
tion may be equal to each other, but, in general, they are entirely
different forces. Consider the example I proposed. The forces
applied to the body (the sled) are the weight P, bearing reaction
N, force F fr of sliding friction and the tension F of the rope
(see Figure 15). We resolve force F into its vertical (Esina) and
horizontal (F cos a) components. All forces acting in the vertical
direction counterbalance one another. This enables us to find the
bearing reaction:

N = P — F sin a (6)

As you can see, this force is not equal to the weight of the sled,
but is less by the amount F sin a. Physically, this is what should
be expected, because the taut rope, being pulled at an angle up¬
wards, seems to “raise” the sled somewhat. This reduces the force
with which the sled bears on the surface and thereby the bearing
reaction as well. So, in this case,

Ff r =k{P—F sin a) (7)

If the rope were horizontal (a = 0), then instead of equation (6)
we would have N = P, from which it follows that Fr r = kP.

STUDENT: I understand now. I never thought about this before.

TEACHER: This is quite a common error of examinees who
attempt to treat the force of sliding friction as the product of
the coefficient of friction by the weight and not by the bearing
reaction. Try to avoid such mistakes in the future.

STUDENT: I shall follow the rule: to find the friction force, first
determine the bearing reaction.

TEACHER: So far we have been dealing with the force of sliding
friction. Now let us consider static friction. This has certain

CAN YOU DETERMINE THE FRICTION FORCE? 33

specific features to which students do not always pay sufficient
attention. Take the following example. A body is at rest on a
horizontal surface and is acted on by a horizontal force F which
tends to move the body. How great do you think the friction
force will be in this case?

STUDENT: If the body rests on a horizontal plane and force F
acts horizontally, then N = P. Is that correct?

TEACHER: Quite correct. Continue.

STUDENT: It follows that the friction force equals kP.

'rr

N

77777777777777,

777777777777777,

Figure 16: Forces acting on a
body at rest.

TEACHER: You have made a typical mistake by confusing the
forces of sliding and static friction. If the body were sliding
along the plane, your answer would be correct. But here the
body is at rest. Hence it is necessary that all forces applied to
the body counterbalance one another. Four forces act on the
body: the weight P, bearing reaction N, force F and the force
of static friction Ff r (Figure 16). The vertical forces P and N
counterbalance each other. So should the horizontal forces F and
Fj r Therefore

F fr =F (8)

STUDENT: It follows that the force of static friction depends on
the external force tending to move the body.

TEACHER: Yes, that is so. The force of static friction increases
with the force F. It does not increase infinitely, however. The

34 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

force of static friction reaches a maximum value:

F max = k 0 N ( 9 )

Coefficient k 0 slightly exceeds coefficient k which characterizes,
according to equation (5), the force of sliding friction. As soon
as the external force F reaches the value k 0 N, the body begins
to slide. At this value, coefficient k 0 becomes equal to k, and so
the friction force is reduced somewhat. Upon further increase
of force F, the friction force (now the force of sliding friction)
ceases to increase further (until very high velocities are attained),
and the body travels with gradually increasing acceleration. The
inability of many examinees to determine the friction force is
disclosed by the following rather simple question: What is the
friction force when a body of weight P is at rest on an inclined
plane with an angle of inclination a? One hears a variety of
incorrect answers. Some say that the friction force equals kP,
and others that it equals kN = kP cos a.

STUDENT: I understand. Since the body is at rest, we have to
deal with the force of static friction. It should be found from
the condition of equilibrium of forces acting along the inclined
plane. There are two such forces in our case: the friction force
F fr and the sliding force P sin a acting downward along the
plane: Therefore, the correct answer is

Ff r = P sin a

TEACHER: Exactly. In conclusion, consider the problem illus¬
trated in Figure 17. A load of mass m lies on a body of mass M;
the maximum force of static friction between the two is charac¬
terized by the coefficient k 0 and there is no friction between the
body and the earth. Find the minimum force F applied to the
body at which the load will begin to slide along it.

STUDENT: First I shall assume that force F is sufficiently small,
so that the load will not slide along the body. Then the two
bodies will acquire the acceleration

M

Figure 17: A load of mass m
lies on a body of mass M. We
have to find the minimum force
F so that load begins to slide.

F

CAN YOU DETERMINE THE FRICTION FORCE? 35

TEACHER: Correct. What force will this acceleration impart to
the load?

STUDENT: It will be subjected to the force of static friction 7y r
by the acceleration. Thus

Ff r = ma =

Fm
M + m

It follows that with an increase in force F, the force of static
friction Fr r also increases. It cannot, however, increase infinitely.
Its maximum value is

= k 0 N = k 0 mg

Consequently, the maximum value of force F. at which the two
bodies can still travel together as an integral unit is determined
from the condition

k 0 mg =

Fm

M + m

This, then, is the minimum force at which the load begins to
slide along the body.

TEACHER: Your solution of the proposed problem is correct. I
am completely satisfied with your reasoning.

§ 4 How Well Do You Know
Newton's Laws Of Motion?

TEACHER: Please state Newton’s first law of motion.

STUDENT: A body remains at rest or in a state of uniform mo¬
tion in a straight line until the action of other bodies compels it
to change that state.

TEACHER: Is this law valid in all frames of reference?

STUDENT: I don’t understand your question.

TEACHER: If you say that a body is at rest, you mean that it is
stationary with respect to some other body which, in the given
case, serves as the reference system, or frame of reference. It
is quite pointless to speak of a body being in a state of rest or
definite motion without indicating the frame of reference. The
nature of the motion of a body depends upon the choice of
the frame of reference. For instance, a body lying on the floor
of a travelling railway car is at rest with respect to a frame of
reference attached to the car, but is moving with respect to a
frame of reference attached to the track. Now we can return
to my question. Is Newton’s first law valid for all frames of
reference?

STUDENT: Well, it probably is.

TEACHER: I see that this question has taken you unawares. Ex¬
periments show that Newton’s first law is not valid for all refer-

38 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

ence systems. Consider the example with the body lying on the
floor of the railway car. We shall neglect the friction between the
body and the floor. First we shall deal with the position of the
body with respect to a frame of reference attached to the car. We
can observe the following: the body rests on the floor and, all of
a sudden, it begins to slide along the floor even though no action
of any kind is evident. Here we have an obvious violation of
Newton’s first law of motion. The conventional explanation of
this effect is that the car, which had been travelling in a straight
line and at uniform velocity, begins to slow down, because the
train is braked, and the body, due to the absence of friction,
continues to maintain its state of uniform straight-line motion
with respect to the railway tracks. From this we can conclude
that Newton’s law holds true in a frame of reference attached to
the railway tracks, but not in one attached to a car being slowed
down.

Frames of reference for which Newton’s first law is valid are
said to be inertial; those in which it is not valid are non-inertial.

For most of the phenomena we deal with we can assume that
any frame of reference is inertial if it is attached to the earth’s
surface, or to any other bodies which are at rest with respect to
the earth’s surface or travel in a straight line at uniform veloc¬
ity. Non-inertial frames of reference are systems travelling with
acceleration (or deceleration), for instance rotating systems, ac¬
celerating or decelerating lifts, etc. Note that not only Newton’s
first law of motion is invalid for non-inertial reference systems,
but his second law as well (since the first law is a particular case
of the second law).

STUDENT: But if Newton’s laws cannot be employed for frames
of reference travelling with acceleration, then how can we deal
with mechanics in such frames?

TEACHER: Newton’s laws of motion can nevertheless be used
for non-inertial frames of reference. To do this, however, it will
be necessary to apply, purely formally, an additional force to the
body. This force, the so called inertial force, equals the product
of the mass of the body by the acceleration of the reference

HOW WELL DO YOU KNOW NEWTON’S LAWS OF MOTION? 39

system, and its direction is opposite to the acceleration of the
body. I should emphasize that no such force actually exists but,
if it is formally introduced, then Newton’s laws of motion will
hold true in a non-inertial frame of reference.

I want to advise you, however, to employ only inertial frames of
reference in solving problems. Then, all the forces that you have
to deal with will be really existing forces.

STUDENT: But if we limit ourselves to inertial frames of refer¬
ence, then we cannot analyse, for instance, a problem about a
body lying on a rotating disk.

TEACHER: Why can’t we? The choice of the frame of reference
is up to you. If in such a problem you use a reference system
attached to the disk (i.e. a non-inertial system), the body is con¬
sidered to be at rest. But if your reference system is attached to
the earth (i.e. an inertial reference system), then the body is dealt
with as one travelling in a circle.

I would advise you to choose an inertial frame of reference. And
now please state Newton’s second law of motion.

STUDENT: This law can be written as F = ma, where F is the
force acting on the body, m is its mass and a - acceleration.

TEACHER: Your laconic answer is very typical. I should make
three critical remarks on your statement; two are not very im¬
portant and one is essential. In the first place, it is not the force
that results from the acceleration, but, on the contrary, the ac¬
celeration is the result of the applied force. It is therefore more
logical to write the equation of the law as

BF

( 10 )

where B is the proportionality factor depending upon the choice
of units of measurement of the quantities in equation (10). No¬
tice that your version had no mention of the proportionality
factor B . Secondly, a body is accelerated by all forces applied to
it (though some may counterbalance one another). Therefore,

40 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

in stating the law you should use, not the term “force”, but the
more accurate term “resultant force”.

My third remark is the most important. Newton’s second law
establishes a relationship between force and acceleration. But
force and acceleration are vector quantities, characterized not
only by their numerical value (magnitude) but by their direction
as well. Your statement of the law fails to specify the directions.
This is an essential shortcoming. Your statement leaves out a
vital part of Newton’s second law of motion. Correctly stated
it is: the acceleration of a body is directly proportional to the
resultant of all forces acting on the body, inversely proportional
to the mass of the body and takes place in the direction of the
resultant force. This statement can be analytically expressed by
the formula

_ BF
a = —
m

( 11 )

(where the arrows over the letters denote vectors).

STUDENT: When in § 2 we discussed the forces applied to a
body thrown upward at an angle to the horizontal, you said you
would show later that the direction of motion of a body does not
necessarily coincide with the direction of the force applied to it.
You referred then to Newton’s second law.

TEACHER: Yes, I remember, and I think it would be quite appro¬
priate to return to this question. Let us recall what acceleration
is. As we know, acceleration is characterized by the change in
velocity in unit time. Illustrated in Figure 18 are the velocity
vectors v x and v 2 of a body for two nearby instants of time t and
t + At. The change in velocity during the time At is the vector
Av = v 2 — By definition, the acceleration is

, ,. Av

a(t)= iim -

V ' At— >o At

( 12 )

or, more rigorously,

-v , ,. Av

a(t)= iim -

At—>o At

Figure 18: Depicting change in
velocity vectorially.

(13)

HOW WELL DO YOU KNOW NEWTON’S LAWS OF MOTION? 41

It follows that the acceleration vector is directed along vector
Av, which represents the change in velocity during a sufficiently
short interval of time. It is evident from Figure 18 that the veloc¬
ity vectors and the change in velocity vector can be oriented in
entirely different directions. This means that, in the general case,
the acceleration and velocity vectors are also differently oriented.
Is that clear?

STUDENT: Yes, now I understand. For example, when a body
travels in a circle, the velocity of the body is directed along a
tangent to the circle, but its acceleration is directed along a radius
toward the centre of rotation (I mean centripetal acceleration).

TEACHER: Your example is quite appropriate. Now let us return
to relationship (equation (11)) and make it clear that it is pre¬
cisely the acceleration and not the velocity that is oriented in the
direction of the applied force, and that it is again the acceleration
and not the velocity that is related to the magnitude of this force.
On the other hand, the nature of a body’s motion at any given
instant is determined by the direction and magnitude of its veloc¬
ity at the given instant (the velocity vector is always tangent to
the path of the body).

Since the acceleration and velocity are different vectors, the
direction of the applied force and the direction of motion of the
body may not coincide in the general case. Consequently, the

nature of the motion of a body at a given instant is not uniquely
determined by the forces acting on the body at the given instant.

STUDENT: This is true for the general case. But, of course, the
direction of the applied force and the velocity may coincide.

TEACHER: Certainly, that is possible. Lift a body and release it
carefully, so that no initial velocity is imparted to it. Here the
direction of motion will coincide with the direction of the force
of gravity. If, however, you impart a horizontal initial velocity to
the body then its direction of motion will not coincide with the
direction of the gravity force; the body will follow a parabolic
path. Though in both cases the body moves due to the action
of the same force - its weight - the nature of its motion differs.

42 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

A physicist would say that this difference is due to the different
initial conditions: at the beginning of the motion the body had
no velocity in the first case and a definite horizontal velocity in
the second.

Illustrated in Figure 19 are the trajectories of bodies thrown with
initial velocities of different directions, but in all cases the same
force, the weight of the body, is acting on it.

STUDENT: Does that mean that the nature of the motion of a
body at a given instant depends not only on the forces acting on
the body at this instant, but also on the initial conditions?

TEACHER: Exactly. It should be emphasized that the initial
conditions reflect the prehistory of the body. They are the result
of forces that existed in the past. These forces no longer exist, but
the result of their action is manifested. From the philosophical
point of view, this demonstrates the relation of the past to the
present, i.e, the principle of causality.

Note that if the formula of Newton’s second law contained the
velocity and not the acceleration, this relationship of the past and
present would not be revealed. In this case, the velocity of a body
at a given instant (i.e. the nature of its motion at a given instant)
would be fully determined by the forces acting on the body
precisely at this instant; the past would have no effect whatsoever
on the present.

I want to cite one more example illustrating the aforesaid. It is
shown in Figure 20: a ball hanging on a string is subject to the
action of two forces, the weight and the tension of the string. If
it is deflected to one side of the equilibrium position and then
released, it will begin to oscillate. If, however, a definite velocity
is imparted to the ball in a direction perpendicular to the plane
of deviation, the ball will begin to travel in a circle at uniform
velocity. As you can see, depending upon the initial conditions,
the ball either oscillates in a plane (see Figure 20 (a)), or travels at
uniform velocity in a circle (see Figure 20 (b)). Only two forces
act on it in either case: its weight and the tension of the string.

Figure 19: Trajectories of
bodies with different initial
velocities.

Figure 20: A ball hanging from
a string can have different
resultant motions depending
on the initial conditions.

HOW WELL DO YOU KNOW NEWTON’S LAWS OF MOTION? 43

STUDENT: I haven’t considered Newton’s laws from this view¬
point.

TEACHER: No wonder then that some students, in trying to de¬
termine the forces applied to a body, base their reasoning on the
nature of motion without first finding out what bodies interact
with the given body. You may recall that you did the same. That
is exactly why, when drawing Figure 8 (c) and Figure 8 (d), it
seemed to you that the sets of forces applied to the body in those
cases should be different. Actually, in both cases two forces are
applied to the body: its weight and the tension of the string.

STUDENT: Now I understand that the same set of forces can
cause motions of different nature and therefore data on the
nature of the motion of a body cannot serve as a starting point in
determining the forces applied to the body.

TEACHER: You have stated the matter very precisely. There is
no need, however, to go to the extremes. Though different kinds
of motion may be caused by the same set of forces (as in Fig¬
ure 20), the numerical relations between the acting forces differ
for the different kinds of motion. This means that there will be
a different resultant applied force for each motion. Thus, for
instance, in uniform motion of a body in a circle, the resultant
force should be the centripetal one; in oscillation in a plane, the
resultant force should be the restoring force. From this it follows
that even though data on the kind of motion of a body cannot
serve as the basis for determining the applied forces, they are far
from superfluous.

In this connection, let us return to the example illustrated in
Figure 20. Assume that the angle a, between the vertical and
the direction of the string is known and so is the weight P of
the body. Find the tension T in the string when (1) the oscil¬
lating body is in its extreme position, and (2) when the body is
travelling uniformly in a circle.

In the first case, the resultant force is the restoring force and it
is perpendicular to the string. Therefore, the weight P of the
body is resolved into two components, with one component

44 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

along the resultant force and the other perpendicular to it (i.e.
directed along the string). Then the forces perpendicular to the
resultant force, i.e. those acting in the direction along the string,
are equated to each other (see Figure 21 (a)). Thus

T l = Pcosa

In the second case, the resultant force is the centripetal one

Figure 21: Resolving the forces
on a moving pendulum.

and is directed horizontally. Hence, the tension T 2 of the string
should be resolved into a vertical and a horizontal force, and the
forces perpendicular to the resultant force, i.e, the vertical forces,
should be equated to each other (Figure 21 (b)).

Then

T 9 cos a = P or T 2 = -

cos a

As you can see, a knowledge of the nature of the body’s motion
proved useful in finding the tension of the string.

STUDENT: If I understand all this correctly, then, knowing the
interaction of bodies, you can find the forces applied to one of
them; if you know these forces and the initial conditions, you
can predict the nature of the motion of the body (the magnitude
and direction of its velocity at any instant). On the other hand,
if you know the kind of motion of a body you can establish the
relationships between the forces applied to it. Am I reasoning
correctly?

HOW WELL DO YOU KNOW NEWTON’S LAWS OF MOTION? 45

TEACHER: Quite so. But let us continue. I want to propose a
comparatively simple problem relating to Newton’s second law
of motion. Two bodies, of masses M and m, are raised to the
same height above the floor and are released simultaneous. Will
the two bodies reach the floor simultaneously if the resistance
of the air is the same for each of them? For simplicity we shall
assume that the air resistance is constant.

STUDENT: Since the air resistance is the same for the two bodies,
it can be disregarded. Consequently, both bodies reach the floor
simultaneously.

TEACHER: You are mistaken. You have no right to disregard the
resistance of the air. Take, for example, the body of mass M. It
is subject to two forces: the weight Mg and the air resistance F.
The resultant force is Mg — F. From this we find the accelera¬
tion. Thus

Mg — F F

a =- = g -

M 6 M

In this manner, the body of larger mass has a higher acceleration
and will, consequently, reach the floor first.

Once more I want to emphasize that in calculating the accel¬
eration of a body it is necessary to take into account I all the
forces applied to it, i.e. you must find the resultant force. In this
connection, the use of the term “driving force” is open to crit¬
icism. This term is inappropriate. In applying it to some force
(or to several forces) we seem to single out the role of this force
(or forces) in imparting acceleration to the body. As if the other
forces concerned were less essential This is absolutely wrong.
The motion of a body is a result of the action of all the forces
applied to it without any exceptions (of course, the initial condi¬
tions should be taken into account).

Let us now consider an example on Newton’s third law of mo¬
tion. A horse starts to pull a waggon. As a result, the horse and
waggon begin to travel with a certain acceleration According to
Newton’s third law, whatever the force with which the horse
pulls the waggon, the waggon pulls back on the horse with ex¬
actly the same force but in the opposite direction. This being so,

46 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

why do the horse and waggon travel forward with an accelera¬
tion? Please explain.

STUDENT: I have never given this any thought but I see no con¬
tradictions. The acceleration would be difficult to ex plain if the
force with which the horse acts on the waggon was counterbal¬
anced by the force with which the waggon acts on the horse. But
these forces cannot cancel each other since they are applied to
different bodies: one to tho horse and the other to the waggon.

Figure 22: Why do the horse
and waggon travel forward
with an acceleration?

TEACHER: Your explanation is applicable to the case when the
waggon is not harnessed to the horse. Then the horse pushes
away from the waggon, as a result of which the waggon moves in
one direction and the horse in the other. The case I proposed is
entirely different. The horse is harnessed to the waggon. Thus
they are linked together and travel as a single system. The forces
of interaction between the horse and waggon that you mentioned
are applied to different parts of the same system. In the motion
of this system as a whole these forces can be regarded as mutually
counterbalancing forces. Thus, you haven’t yet answered my
question.

STUDENT: Well, then I can’t understand what the matter is.
Maybe the action here is not fully counterbalanced by the reac¬
tion? After all a horse is a living organism.

TEACHER: Now don’t let your imagination run away with you.

It was sufficient for you to meet with some difficulty and you
are ready to sacrifice one of the principal laws of mechanics. To
answer my question, there is no need to revise Newton’s third
law of motion. On the contrary, let us use this law as a basis for
our discussion.

According to the third law. the interaction of the horse and the
waggon cannot lead to the motion of this system as a whole (or,

jr

*0 •

^ 0 — 0 -

W777777ZW7?,

V////////////////////////7

?

HOW WELL DO YOU KNOW NEWTON’S LAWS OF MOTION? 47

more precisely, it cannot impart acceleration to the system as a
whole). This being so there must exist some kind of supplemen¬
tary interaction. In other words at least one more body must
participate in the problem in addition to the horse and waggon.
This body, in the given case is the earth. As a result, we have
three interactions to deal’ with instead of one, namely:

(1) between the horse and the waggon (we shall denote this
force by f 0 ;

(2) between the horse and the earth (force F), in which the
horse pushes against the ground; and

(3) between the waggon and the earth (force /) which is the
friction of the waggon against the ground.

All bodies are shown in Figure 22: the horse, the waggon and the
earth and two forces are applied to each body. These two forces
are the result of the interaction of the given body with the two
others. The acceleration of the horse-waggon system is caused
by the resultant of all the forces applied to it. There are four
such forces and their resultant is F — f. This is what causes the
acceleration of the system. Now you see that this acceleration
is not associated with the interaction between the horse and the
waggon.

STUDENT: So the earth’s surface turns out to be, not simply the
place on which certain events occur, but an active participant of
these events.

TEACHER: Your pictorial comment is quite true. Incidentally, if
you locate the horse and waggon on an ideal icy surface, thereby
excluding all horizontal interaction between this system and
the earth, there will be no motion, whatsoever. It should be
stressed that no internal interaction can impart acceleration to a
system as a whole. This can be done only by external action (you
can’t lift yourself by your hair, or bootstraps either). This is an
important practical inference of Newton’s third law of motion.

If you know mechanics well, you can easily solve problems. The con¬
verse is just as true: if you solve problems readily, you evidently have a
good knowledge of mechanics. Therefore, extend your knowledge of
mechanics by solving as many problems as you can.

§ 5 How Do You Go About Solving Problems In
Kinematics?

TEACHER: Assume that two bodies are falling from a certain
height. One has no initial velocity and the other has a certain
initial velocity in a horizontal direction. Here and further on we
shall disregard the resistance of the air. Compare the time it takes
for the two bodies to fall to the ground.

STUDENT: The motion of a body thrown horizontally can be re¬
garded as a combination of two motions: vertical and horizontal.
The time of flight is determined by the vertical component of the
motion. Since the vertical motions of the bodies are determined
in both cases by the same data (same height and the absence of a
vertical component of the initial velocity), the time of fall is the
same for the two bodies. It equals yjlH/ g, where H is the initial
height.

TEACHER: Absolutely right. Now let us consider a more com¬
plex case. Assume that both bodies are falling from the height H
with no initial velocity, but in its path one of them meets a fixed
plane, inclined at an angle of 45° to the horizontal. As a result
of this impact on the plane the direction of the velocity of the
body becomes horizontal (Figure 23). The point of impact is at
the height h. Compare the times of fall of the two bodies.

STUDENT: Both bodies take the same time to fall to the level of
the inclined plane. As a result of the impact on the, plane one

52

QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 23: Two bodies falling
from height H . One of them
meets a fixed plane at height h.
Problem is to compare the time
of fall of the two bodies.

of the bodies acquires a horizontal component of velocity. This
horizontal component cannot, however, influence the vertical
component of the body’s motion. Therefore, it follows that in
this case as well the time of fall should be the same for the bodies.

TEACHER: Your answer is wrong. You were right in saying that
the horizontal component of the velocity doesn’t influence the
vertical motion of the body and, consequently, its time of fall.
When the body strikes the inclined plane it not only acquires
a horizontal velocity component, but also loses the vertical
component of its velocity, and this of course must affect the time
of fall After striking the inclined plane, the body falls from the
height h with no initial vertical velocity. The impact against
the plane delays the vertical motion of the body and thereby
increases its time of fall. The time of fall for the body which
dropped straight to the ground is y^2 H/g; that for the body
striking the plane is ^2(H — h)/g + ^Jlh/g.

This leads us to the following question: at what h to H ratio will
the time of fall reach its maximum value? In other words, at what
height should the inclined plane be located so that it delays the
fall most effectively?

STUDENT: I am at a loss to give you an exact answer. It seems to
me that the ratio h/H should not be near to 1 or to 0, because
a ratio of 1 or 0 is equivalent to the absence of any plane what¬
soever. The inclined plane should be located somewhere in the
middle between the ground and the initial point.

HOW DO YOU GO ABOUT SOLVING PROBLEMS IN KINEMATICS? 53

TEACHER: Your qualitative remarks are quite true. But you
should find no difficulty in obtaining the exact answer. We can
write the time of fall of the body as

t =

Now we find the value of x at which the function t (x) is a maxi¬
mum. First we square the time of fall. Thus

t 2 = — ^1 + 2^/(1 — x) x ^

If the time is maximal, its square is also maximal. It is evident
from the last equation that f 2 is a maximum when the function
y = (1 — x) x is a maximum. Thus, the problem is reduced to
finding the maximum of the quadratic trinomial

y = —x + x = —

This trinomial is maximal at x = -. Thus, height h should be
one half of height H.

Our further discussion on typical procedure for solving prob¬
lems in kinematics will centre around the example of a body
thrown upward at an angle to the horizontal (usually called the
elevation angle).

STUDENT: I’m not very good at such problems.

TEACHER: We shall begin with the usual formulation of the
problem: a body is thrown upward at an angle of or, to the hori¬
zon with an initial velocity of v 0 . Find the time of flight T ,
maximum height reached H and the range L.

As usual, we first find the forces acting on the body. The only
force is gravity. Consequently, the body travels at uniform ve¬
locity in the horizontal direction and with uniform acceleration
g in the vertical direction. We are going to deal with the verti¬
cal and horizontal components of motion separately, for which

54 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

purpose we resolve the initial velocity vector into the vertical
(v 0 sin a,) and horizontal (v 0 cos a,) components. The horizon¬
tal velocity component remains constant throughout the flight
while the vertical component varies as shown in Figure 24. Let

Figure 24: A body thrown with
an angle a to the horizon. The
problem is to find the time
of flight, maximum height
reached and the range of the
body.

us examine the vertical component of the motion. The time of
flight T = T x + T 2 , where 7) is the time of ascent (the body
travels vertically with uniformly decelerated motion) and T 2 is
the time of descent (the body travels vertically downward with
uniformly accelerated motion). The vertical velocity of the body
at the highest point of its trajectory (at the instant t = 7)) is
obviously equal to zero. On the other hand, this velocity can be
expressed by the formula showing the dependence of the velocity
of uniformly decelerated motion on time. Thus we obtain

0 = v^sma — gTj

or

7*i =

v 0 sin a
g

(14)

When 7j is known we can obtain

„ t • gfi 2 v 2 sm 2 a
H = v 0 T t sin a — = —--

2 2 g

(15)

The time of descent T 2 can be calculated as the time a body falls
from the known height 77 without any initial vertical velocity:

HOW DO YOU GO ABOUT SOLVING PROBLEMS IN KINEMATICS? 55

Comparing this with equation (14) we see that the time of de¬
scent is equal to the time of ascent. The total time of flight is

„ „ „ 2 t; n sin a

t = t x + t 2 = — -

g

(16)

To find the range L, or horizontal distance travelled, we make
use of the horizontal component of motion. As mentioned
before, the body travels horizontally at uniform velocity. Thus

vl sin 2 a

L=(v 0 cosa)T= - (17)

g

It can be seen from equation (17) that if the sum of the angles at
which two bodies are thrown is 90° and if the initial velocities
are equal, the bodies will fall at the same point. Is everything
clear to you so far?

STUDENT: Why yes, everything seems to be clear.

TEACHER: Fine. Then we shall add some complications. Assume
that a horizontal tail wind of constant force F acts on the body.
The weight of the body is P. Find, as in the preceding case, the
time of flight T, maximum height reached H, and range L.

STUDENT: In contrast to the preceding problem, the horizontal
motion of the body is not uniform; now it travels with a hori¬
zontal acceleration of a = ( F/P) g.

TEACHER: Have there been any changes in the vertical compo¬
nent of motion?

STUDENT: Since the force of the wind acts horizontally the wind
cannot affect the vertical motion of the body.

TEACHER: Good. Now tell me which of the sought for quantities
should have the same values as in the preceding problem.

STUDENT: These will evidently be the time of flight T and the
height FI. They are the ones determined on the basis of the
vertical motion of the body. They will therefore be the same as
in the preceding problem.

56 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: Excellent. How about the range?

STUDENT: The horizontal acceleration and time of flight being
known, the range can be readily found. Thus

aT 2 v* sin 2a IF v\sm 2 a

L ~ \ v o cos cc) T + — =-+ —-

2 g p g

TEACHER: Quite correct. Only the answer would best be written
in another form:

L =

fn sin2 a

g

F

1 H-tan a

P

(18)

Next we shall consider a new problem: a body is thrown at an
angle a to an inclined plane which makes the angle /3 with the
horizontal (Figure 25). The body’s initial velocity is v 0 . Find the
distance L from the point where the body is thrown to the point
where it falls on the plane.

STUDENT: I once made an attempt to solve such a problem but
failed.

TEACHER: Can’t you see any similarity between this problem
and the preceding one?

STUDENT: No, I can’t.

TEACHER: Fet us imagine that the figure for this problem is
turned through the angle j3 so that the inclined plane becomes
horizontal (Figure 26 (a)).

Then the force of gravity is no longer vertical. Now we resolve
it into a vertical {P cos /3) and a horizontal (Tsin p) component.

Figure 25: A body thrown
with an angle a to an inclined
plane. The problem is to find
the distance at which the body
will land on the inclined plane.

HOW DO YOU GO ABOUT SOLVING PROBLEMS IN KINEMATICS? 57

Figure 26: Rotating the Fig¬
ure 25 through angle [3 gives
us a problem similar to the one
we have solved in the previous
section.

P sin /3 for F, P cos j3 for P, and g cos /3 for g
Then we obtain

vi sin 2a

L= -— (1+tan/3tana) (19)

g cos p

At /? = 0, this coincides with equation (17).

Of interest is another method of solving the same problem. We
introduce the coordinate axes Ox and Oy with the origin at
the point the body is thrown from (Figure 26 (b)). The inclined
plane is represented in these coordinates by the linear function

y 1 = — xtan j3

and the trajectory of the body is described by the parabola
y 2 = ax 2 + bx

in which the factors a and b can be expressed in terms of v 0 , a
and j3. Next we find the coordinate x A of the point A of intersec¬
tion of functions y 1 and y 2 by equating the expressions for these

You can readily see now that we have the preceding problem
again, in which the force P sin j3 plays the role of the force of the
wind, and P cos [ 3 the role of the force of gravity. Therefore we
can find the answer by making use of equation (18) provided that
we make the following substitutions:

58 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

functions

—xtan j3 = ax 2 + i

From this it follows that

x A =

tan (3+b

—a

Then we can easily find the required distance

x A tan /3 + b

cos j3 a cos f3

( 20 )

It remains to express factors a and b in terms of v 0 , a and f3.

For this purpose, we examine two points of the parabola - B and
C (see Figure 26 (b)). We write the equation of the parabola for
each of these points:

yic = ax l + bx c )
y 1B = axj + bx B J

The coordinates of points C and B are known to us. Conse¬
quently, the preceding system of equations enables us to de¬
termine factors a and b. I suggest that in your spare time you
complete the solution of this problem and obtain the answer in
the form of equation (19).

STUDENT: I like the first solution better.

TEACHER: That is a matter of taste. The two methods of solution
differ essentially in their nature. The first could be called the
“physical” method. It employs simulation which is so typical of
the physical approach (we slightly altered the point of view and
reduced our problem to the previously discussed problem with
the tail wind). The second method could be called “mathemati¬
cal”. Here we employed two functions and found the coordinates
of their points of intersection.

In my opinion, the first method is the more elegant, but less
general. The field of application of the second method is substan¬
tially wider. It can, for instance, be applied in principle when the
profile of the hill from which the body is thrown is not a straight

HOW DO YOU GO ABOUT SOLVING PROBLEMS IN KINEMATICS? 59

line. Here, instead of the linear function y lt some other func¬
tion will be used which conforms to the profile of the hill. The
first method is inapplicable in principle in such cases. We may
note that the more extensive field of application of mathematical
methods is due to their more abstract nature.

Problems

1. Body A is thrown vertically upward with a velocity of 20 m/s. At what
height was body B which, when thrown at a horizontal velocity of 4 m/s
at the same time body A was thrown, collided with it in its flight? The
horizontal distance between the initial points of the flight equals 4 m. Find
also the time of flight of each body before the collision and the velocity of
each at the instant of collision.

2. From points A and B, at the respective heights of 2 m and 6 m, two bodies
are thrown simultaneously towards each other: one is thrown horizontally
with a velocity of 8 m/s and the other, downward at an angle of 45° to the
horizontal and at an initial velocity such that the bodies collide in flight.
The horizontal distance between points A and B equals 8 m. Calculate the
initial velocity v Q of the body thrown at an angle of 45°, the coordinates

x and y of the point of collision, the time of flight t of the bodies before
colliding and the velocities V/y and v B of the two bodies at the instant of
collision. The trajectories of the bodies lie in a single plane.

3. Two bodies are thrown from a single point at the angles and a 2 to the
horizontal and at the initial velocities tq and v 2 , respectively. At what
distance from each other will the bodies be after the time f ? Consider two
cases:

(1) the trajectories of the two bodies lie in a single plane and the bodies
are thrown in opposite directions, and

(2) the trajectories lie in mutually perpendicular planes.

4. A body falls from the height H with no initial velocity. At the height h it
elastically bounces off a plane inclined at an angle of 30° to the horizontal.
Find the time it takes the body to reach the ground.

5. At what angle to the horizontal (elevation angle) should a body of weight
P be thrown so that the maximum height reached is equal to the range?
Assume that a horizontal tail wind of constant force F acts on the body in
its flight.

60 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

6. A stone is thrown upward, perpendicular to an inclined plane with an angle
of inclination a. If the initial velocity is v Q , at what distance from the point
from which it is thrown will the stone fall?

7. A boy 1.5 m tall, standing at a distance of 15 m from a fence 5 m high,
throws a stone at an angle of 45° to the horizontal. With what minimum
velocity should the stone be thrown to fly over the fence?

§ 6 How Do You Go About
Solving Problems In Dynamics?

TEACHER: In solving problems in dynamics it is especially im¬
portant to be able to determine correctly the forces applied to the
body (see § 2).

STUDENT: Before we go any further, I wish to ask one question.
Assuming that I have correctly found all the forces applied to the
body, what should I do next?

TEACHER: If the forces are not directed along a single straight
line, they should be resolved in two mutually perpendicular
directions. The force components should be dealt with separately
for each of these directions, which we shall call “directions of
resolution”.

We can begin with some practical advice. In the first place, the
forces should be drawn in large scale to avoid confusion in resolv¬
ing them. In trying to save space students usually represent forces
in the form of almost microscopic arrows, and this does not help.
You will understand what I mean if you compare your drawing
(Figure 8) with mine (Figure 9). Secondly, do not hurry to re¬
solve the forces before it can be done properly. First you should
find all forces applied to the body, and show them in the draw¬
ing. Only then can you begin to resolve some of them. Thirdly,
you must remember that after you have resolved a force you
should “forget” about its existence and use only its components.
Either the force itself, or its components, no compromise.

62 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT: How do I choose the directions of resolution?

TEACHER: In making your choice you should consider the nature
of the motion of the body. There are two alternatives: (1) the
body is at rest or travels with uniform velocity in a straight line,
and (2) the body travels with acceleration and the direction of
acceleration is given (at least its sign).

In the first case you can select the directions of resolution arbi¬
trarily, basing (or not basing) your choice on considerations of
practical convenience. Assume, for instance, that in the case illus¬
trated in Figure 10 the body slides with uniform velocity up the
inclined plane. Here the directions of resolution may be (with
equal advantage) either vertical and horizontal (Figure 27 (a)) or
along the inclined plane and perpendicular to it (Figure 27 (b)).

(a)

(b)

Figure 27: Two ways of
resolving the forces applied to a
body in motion.

After the forces have been resolved, the algebraic sums of the
component forces for each direction of resolution are equated to
zero (remember that we are still dealing with the motion of bod¬
ies without acceleration). For the case illustrated in Figure 27 (a)
we can write the system of equations

N cos a — Fr r sin a — P = 0
F — iy r cos a — Asin a = 0

(21)

HOW DO YOU GO ABOUT SOLVING PROBLEMS IN DYNAMICS? 63

The system of equations for the case in Figure 27 (b) is

N — P cos a — F sin a = 0
Fjr r + P sin a — F cos a = 0

( 22 )

STUDENT: But these systems of equations differ from each other.

TEACHER: They do but, nevertheless, lead to the same results, as
can readily be shown. Suppose it is required to find the force F
that will ensure the motion of the body at uniform velocity up
along the inclined plane. Substituting equation (5) into equations
(21) we obtain

N (cos a — k sin a) — P = 0 |

F —N {k cos a + sin a) = 0 J

From the first equation of this system we get

N =

P

cos a — k sin a

which is substituted into the second equation to determine the
required force. Thus

F = P

k cos a + sin a

cos a

:sin a

Exactly the same answer is obtained from equations (22). You
can check this for yourself.

STUDENT: What do we do if the body travels with acceleration?

TEACHER: In this case the choice of the directions of resolution
depends on the direction in which the body is being accelerated
(direction of the resultant force). Forces should be resolved in
a direction along the acceleration and in one perpendicular to
it The algebraic sum of the force components in the direction
perpendicular to the acceleration is equated to zero, while that
of the force components in the direction along the acceleration
is equal, according to Newton’s second law of motion, to the
product of the mass of the body by its acceleration.

64 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Let us return to the body on the inclined plane in the last prob¬
lem and assume that the body slides with a certain acceleration
up the plane. According to my previous remarks, the forces
should be resolved as in the case shown in Figure 27 (b). Then, in
place of equations (22), we can write the following system

N — P cos a — F sin a = 0

(23)

F cos a — Fr r — Psina = ma = P

Making use of equation (5), we find the acceleration of the body

STUDENT: In problems of this kind dealing with acceleration,
can the forces be resolved in directions other than along the
acceleration and perpendicular to it? As far as I understand from
your explanation, this should not be done.

TEACHER: Your question shows that I should clear up some
points. Of course, even in problems on acceleration you have a
right to resolve the forces in any two mutually perpendicular di¬
rections. In this case, however, you will have to resolve not only
the forces, but the acceleration vector as well. This method of
solution will lead to additional difficulties. To avoid unnecessary
complications, it is best to proceed exactly as is advised. This is
the simplest course. The direction of the body’s acceleration is
always known (at least its sign), so you can proceed on the basis
of this direction. The inability of examinees to choose the di¬
rections of force resolution rationally is one of the reasons for
their helplessness in solving more or less complex problems in
dynamics.

STUDENT: We have only been speaking about resolution in
two directions. In the general case, however, it would probably
be more reasonable to speak of resolution in three mutually
perpendicular directions. Space is actually three-dimensional.

TEACHER: You are absolutely right. The two directions in
our discussions are explained by the fact that we are dealing

HOW DO YOU GO ABOUT SOLVING PROBLEMS IN DYNAMICS? 65

with plane (two-dimensional) problems. In the general case,
forces should be resolved in three directions. All the remarks
made above still hold true, however. I should mention that, as
a rule, two-dimensional problems are given in examinations.
Though, of course, the examinee may be asked to make a not-
too-complicated generalization for the three-dimensional case.

Problems

8. A body with a mass of 5 kg is pulled along a horizontal plane by a force
of 3 N 1 applied to the body at an angle of 30° to the horizontal. The
coefficient of sliding friction is 0.2. Find the velocity of the body 10 s after
the pulling force begins to act, and the work done by the friction force
during this time.

9. A man pulls two sleds tied together by applying a force of F = 12 N to the
pulling rope at an angle of 45° to the horizontal (Figure 28). The masses of
the sleds are equal to = m-, = 15 kg. The coefficient of friction between
the runners and the snow is 0.02. Find the acceleration of the sleds, the
tension of the rope tying the sleds together, and the force with which the
man should pull the rope to impart uniform velocity to the sleds.

1 Originally unit of kgf is
used, here after we will use the
unit of Newtons denoted by N.

Figure 28: Sled being pulled by
a rope with an angle 45°. See
Problem 9.

10. Three equal weights of a mass of 2 kg each are hanging on a string passing
over a fixed pulley as shown in Figure 29. Find the acceleration of the
system and the tension of the string connecting weights 1 and 2.

V/////J///.

Figure 29: A system of three in¬
teracting masses. See Problem
10 .

1

66

QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

11. Calculate the acceleration of the weights and the tension in the strings for
the case illustrated in Figure 30. Given: a = 30°, P l = 4N, P 2 = 2newton,
and P 3 = 8 N. Neglect the friction between the weights and the inclined
plane.

12. Consider the system of weights shown in Figure 31. Here P\ = 1 NP 2 =
2N, P$ = 8N and P4 — 0.5 N, and a = 30°. The coefficient of friction
between the weights and the planes equals 0.2. Find the acceleration of the
set of weights, the tension of the strings and the force with which weight
presses downward on weight P 3 .

Figure 30: A system of three
masses on an incline. See
problem 11.

P,

Figure 31: A system of three
masses on an incline. See
problem 12

% 7 Are Problems In Dynamics Much More Diffi¬
cult To Solve If Friction Is Taken Into Account?

TEACHER: Problems may become much more difficult when the
friction forces are taken into account.

STUDENT: But we have already discussed the force of friction
(§ 3). If a body is in motion, the friction force is determined
from the bearing reaction (iy r = kN); if the body is at rest, the
friction force is equal to the force that tends to take it out of this
state of rest. All this can readily be understood and remembered.

TEACHER: That is so. However, you overlook one important
fact. You assume that you already know the answers to the
following questions: (1) Is the body moving or is it at rest? (2) In
which direction is the body moving (if at all)?

If these items are known beforehand, then the problem is com¬
paratively simple. Otherwise, it may be very complicated from
the outset and may even require special investigation.

STUDENT: Yes, now I recall that we spoke of this matter in § 2
in connection with our discussion concerning the choice of the
direction of the friction force.

TEACHER: Now I want to discuss this question in more detail. It
is my firm opinion that the difficulties involved in, solving prob¬
lems which take the friction force into account are obviously

68 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

underestimated both by students and by certain authors who
think up problems for physics textbooks.

Let us consider the example illustrated in Figure 10. The angle
of inclination a of the plane, weight P of the body, force F and
the coefficient of friction k are given. For simplicity we shall
assume that k 0 = k (where k 0 is the coefficient determining
the maximum possible force of static friction). It is required
to determine the kind of motion of the body and to find the
acceleration.

Let us assume that the body slides upward along the inclined
surface. We can resolve the forces as shown in Figure 27 (b) and
make use of the result obtained for the acceleration in § 6.Thus

g

a = —[F cos a — P sin a — (P cos a + F sin a) k\ (24)

It follows from equation (24) that for the body to slide upward
along the inclined plane, the following condition must be com¬
plied with:

F cos a — P sin a — (P cos a + F sin a) k ^ 0

This condition can be written in the form

or

F^P

k cos a 4- sin a
cos a — ksma

F^P

k + tan a
1 — kx.-s.na

(25)

We shall also assume that the angle of inclination of the plane is
not too large, so that {\ — k tan a) > 0 or

1

tan a < — (26)

k

We shall next assume that the body slides downward along the
inclined plane. We again resolve all the forces as in Figure 27 (b)
but reverse the friction force. As a result we obtain the fall owing
expression for the acceleration of the body

a = — [P sin a — F cos a — (P cos a + F sin a) k\ (27)

ARE PROBLEMS IN DYNAMICS MUCH MORE DIFFICULT TO SOLVE IF FRICTION IS TAKEN INTO

ACCOUNT? 69

From equation (27) it follows that for the body to slide down¬
ward along the inclined plane, the following condition must be
met:

P sin a — F cos a — (P cos a + F sin a) k 3 s 0
This condition we write in the form

F ^P

: + ki

or

„ „ f tan a — k\

F^P ---

V 1 + k tun a J

(28)

In this case, we shall assume that the angle of inclination of the
plane is not too small, so that (tan a — k) > 0, or

tan a > k

(29)

Combining conditions (25), (26), (28) and (29), we can come to
the following conclusions:

(1) Assume that the condition

, 1

k < tan a < —

holds good for an inclined plane. Then:
k + tan a

(a) if F > P

, the body slides upward

1 — k tan a ,

with an acceleration that can be determined by
equation (24);

(b) if F = P ( -—i^, the body slides upward at

1 —ktzn a
uniform velocity or is at rest;
tan a — k

(c) if F < P

, the body slides downward

, 1 + k tan a,

with an acceleration that can be determined by
equation (27);

(d) if F = P ( - - - -'j, the body slides downward

(e) if P

\ + k tan a
with uniform velocity or is at rest;

tan a — k \ „ „ / k + tan a

1 + k tan a
body is at rest.

< F < P

1 — k tan a

, the

70 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Note that upon increase in force F from P

tan a — k
1 + k tana

to P ( —-t--^, the force of static friction is gradually

V1 — k tan a )

reduced from k (P cos a + F sin a) to zero; then, after its
direction is reversed, it increases to the value k (P cos a +

F sin a). While this goes on the body remains at rest.

(2) Now assume that the inclined plane satisfies the condition

0 < tana ^ k

then:

(a) if F > P

k + tan a

, the body slides upward

1 — ktana y

with an acceleration that can be determined by
equation (24);

(b) if F = P ( - - - -^, the body slides upward at

,1 — &tan a,
uniform velocity or is at rest;

.. / k 4- tana \ .

(c) it F < P [ -- - - ] , the body is at rest; no

tan a,

downward motion of the body along the inclined
plane is possible (even if force F vanishes).

(3) Finally, let us assume that the inclined plane meets the
condition

1

tan a ^ —

k

then:

(a) if F < P

tan a — k

, the body slides downward

\ + k tan a,

with an acceleration that can be determined by
equation (27);

tana — k

(b) if F = P

, the body slides down ward

. 1 + k tan a
with uniform velocity or is at rest;

( tana — k \ . . .

(c) it F > P [ --- , the body is at rest; no

\ 1 + k tan a J

upward motion of the body along the inclined
plane is possible. On the face of it, this seems
incomprehensible because force F can be increased
indefinitely! The inclination of the plane is so

ARE PROBLEMS IN DYNAMICS MUCH MORE DIFFICULT TO SOLVE IF FRICTION IS TAKEN INTO

ACCOUNT? 71

large, however, that, with an increase in force F, the
pressure of the body against the plane will increase
at an even faster rate.

STUDENT: Nothing of the kind has ever been demonstrated to us
in school.

TEACHER: That is exactly why I wanted to draw your attention
to this matter. Of course, in your entrance examinations you will
evidently have to deal with much simpler cases: there will be no
friction, or there will be friction but the nature of the motion
will be known beforehand (for instance, whether the body is in
motion or at rest). However, even if one does not have to swim
over deep spots, it is good to know where they are.

STUDENT: What will happen if we assume that k = 0?

TEACHER: In the absence of friction, everything becomes much
simpler at once. For any angle of inclination of the plane, the
results will be:

► at F > P tan a, the body slides upward with the accelera¬
tion

a = (i 7 cos or — Psina) (30)

► at F = P tan a, the body slides with uniform velocity
(upward or downward) or is at rest;

► at F < P tan a, the body slides downward with an accelera¬
tion

a = ^ (P sin a — F cos a) (31)

Note that the results of equations (30) and (31) coincide with an
accuracy to the sign. Therefore, in solving problems, you can

safely assume any direction of motion, find a and take notice
of the sign of the acceleration. If a >0, the body travels in the
direction you have assumed; if a < 0, the body will travel in the
opposite direction (the acceleration will be equal to \a\).

Let us consider one more problem. Two bodies and P 2 are
connected by a string running over a pulley. Body Pj is on an

72 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

inclined plane with the angle of inclination a and coefficient
of friction k; body P 2 hangs on the string (Figure 32). Find the
acceleration of the system.

Figure 32: Two bodies con¬
nected via string on an inclined
plane with friction. The prob¬
lem is to find the acceleration
of the system.

Assume that the system is moving from left to right. Consid¬
ering the motion of the system as a whole, we can write the
following equation for the acceleration:

: = g

P 2 — Pi sin a — P cos a

P l+ P 2

Assuming now that the system moves from right to left, we
obtain

(32)

: = g

Pi sin a — P 2 — P x k cos a

Pi+P 2

We will carry out an investigation for the given a, and k values,
varying the ratio p = P 2 /Pi- From equation (32) it follows that
for motion from left to right, the condition

(33)

1

P^~.—TI -

sin a + k cos a

should be met. Equation (33) implies that for motion from right
to left the necessary condition is

1

P>~. - 1 -

sin a — k cos a

Here an additional condition is required: the angle of inclination
should not be too small, i.e. tana' > k. If tana: ^ k, then the
system will not move from right to left, however large the ratio
p may be.

If tana: > k, the body is at rest provided the following inequality
holds true:

1 1

<p<

sin a + k cos a

sin a — k cos a

ARE PROBLEMS IN DYNAMICS MUCH MORE DIFFICULT TO SOLVE IF FRICTION IS TAKEN INTO

ACCOUNT? 73

If, instead, tana ^ k, then the body is at rest at

1

P> - —T -

sin a + k cos a

STUDENT: And what will happen if we change the angle a or the
coefficient k ?

TEACHER: I leave investigation from this point of view to you as
a home assignment (see Problems Nos. 13 and 14).

Problems

13. Investigate the problem illustrated in Figure 32 assuming that the angle a
of inclination of the plane and the ratio p = P 2 /P 1 are given, and assigning
various values to the coefficient k.

14. Investigate the problem illustrated in Figure 32, assuming that the coeffi¬
cient of friction k and the ratio p = P 2 /P 1 are given and assigning various
values to the angle a of inclination of the plane. For the sake of simplicity,
use only two values of the ratio: p = 1 (the bodies are of equal weight)
and p = 1/2 (the body on the inclined plane is twice as heavy as the one
suspended on the string).

Motion in a circle is the simplest form of curvilinear motion. The more
important it is to comprehend the specific features of such motion. You
can see that the whole universe is made up of curvilinear motion. Let us
consider uniform and non-uniform motion of a material point in a circle,
and the motion of orbiting satellites. This will lead us to a discussion of
the physical causes of the weightlessness of bodies.

§ 8 How Do You Deal With Motion In A Circle?

TEACHER: I have found from experience that questions and
problems concerning motion of a body in a circle turn out to be
extremely difficult for many examinees. Their answers to such
questions contain a great many typical errors. To demonstrate
this let us invite another student to take part in our discussion.
This student doesn’t know what we have discussed previously.
We shall conditionally call him “STUDENT B” (the first student
will hereafter be called “STUDENT A”).

Will STUDENT B please indicate the forces acting on a satellite,
or sputnik, in orbit around the earth? We will agree to neglect
the resistance of the atmosphere and the attraction of the moon,
sun and other celestial bodies.

STUDENT B: The satellite is subject to two forces: the attraction
of the earth and the centrifugal force.

TEACHER: I have no objections to the attraction of the earth,
but I don’t understand where you got the centrifugal force from.
Please explain.

STUDENT B: If there were no such force, the satellite could not
stay in orbit.

TEACHER: And what would happen to it?

STUDENT B: Why, it would fall to the earth.

78 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: (turning to STUDENT A): Remember what I told
you before! This is a perfect example of an attempt to prove
that a certain force exists, not on the basis of the interaction of
bodies, but by a backdoor manoeuvre - from the nature of the
motion of bodies. As you see, the satellite must stay in orbit,
so it is necessary to introduce a retaining force. Incidentally, if
this centrifugal force really did exist, then the satellite could not
remain in orbit because the forces acting on the satellite would
cancel out and it would fly at uniform velocity and in a straight
line.

STUDENT A: The centrifugal force is never applied to a rotating
body. It is applied to the tie (string or another bonding member).
It is the centripetal force that is applied to the rotating body.

STUDENT B: Do you mean that only the weight is applied to the
satellite?

TEACHER: Yes, only its weight.

STUDENT B: And, nevertheless, it doesn’t fall to the earth?

TEACHER: The motion of a body subject to the force of gravity is
called falling. Hence, the satellite is falling. However, its “falling”
is in the form of motion in a circle around the earth and there¬
fore can continue indefinitely. We have already established that
the direction of motion of a body and the forces acting on it do
not necessarily coincide (see § 4).

STUDENT B: In speaking of the attraction of the earth and the
centrifugal force, I based my statement on the formula

GmM mv 1

where the left-hand side is the force of attraction (w=mass of
the satellite, M =mass of the earth, r=radius of the orbit and
G=gravitational constant), and the right-hand side is the cen¬
trifugal force (f=velocity of the satellite). Do you mean to say
that this formula is incorrect?

HOW DO YOU DEAL WITH MOTION IN A CIRCLE? 79

TEACHER: No, the formula is quite correct. What is incorrect
is your interpretation of the formula. You regard equation (34)
as one of equilibrium between two forces. Actually, it is an
expression of Newton’s second law of motion

where F

GmM

--— and <2

r 2

F = ma

(34a)

is the centripetal acceleration.

STUDENT B: I agree that your interpretation enables us to get
along without any centrifugal force. But, if there is no centrifugal
force, there must at least be a centripetal force. You have not,
however, mentioned such a force.

TEACHER: In our case, the centripetal force is the force of attrac¬
tion between the satellite and the earth. I want to underline the
fact that this does not refer to two different forces. By no means.
This is one and the same force.

STUDENT B: Then why introduce the concept of a centripetal
force at all?

TEACHER: I fully agree with you on this point. The term “cen¬
tripetal force”, in my opinion, leads to nothing but confusion.
What is understood to be the centripetal force is not at all an in¬
dependent force applied to a body along with other forces. It is,
instead, the resultant of all the forces applied to a body travelling
in a circle at uniform velocity.

The quantity mv 2 /r is not a force. It represents the product of
the mass m. of the body by the centripetal acceleration v 2 /r.

This acceleration is directed toward the centre and, consequently,
the resultant of all forces, applied to a body travelling in a circle
at uniform velocity, is directed toward the centre. Thus, there
is a centripetal acceleration and there are forces which, added
together, impart a centripetal acceleration to the body.

STUDENT B: I must admit that this approach to the motion of a
body in a circle is to my liking. Indeed, this motion is not a static
case, for which an equilibrium of forces is characteristic, but a
dynamic case.

80 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT A: If we reject the concept of a centripetal force, then
we should probably drop the term “centrifugal force” as well,
even in reference to ties.

TEACHER: The introduction of the term “centrifugal force” is
even less justified. The centripetal force actually exists, if only
as a resultant force. The centrifugal force does not even exist in
many cases.

STUDENT A: I don’t understand your last remark. The centrifu¬
gal force is introduced as a reaction to the centripetal force. If
it does not always exist, as you say, then Newton’s third law of
motion is not always valid. Is that so?

TEACHER: Newton’s third law is valid only for real forces deter¬
mined by the interaction of bodies, and not for the resultants of
these forces. I can demonstrate this by the example of the conical
pendulum that you are already familiar with (Figure 33).

Figure 33: A conical pendulum.
The body attached to the string
performs circular motion in the
horizontal plane. During this
motion the solid that it traces
has the shape of a cone.

The ball is subject to two forces: the weight P and the tension
T of the string. These forces, taken together, provide the cen¬
tripetal acceleration of the ball, and their sum is called the cen¬
tripetal force. Force P is due to the interaction of the ball with
the earth. The reaction of this force is force P l which is applied
to the earth. Force T results from interaction between the ball
and the string. The reaction of this force is force 7) which is
applied to the string.

If forces P] and 7) are formally added together we obtain a force
which is conventionally understood to be the centrifugal force

HOW DO YOU DEAL WITH MOTION IN A CIRCLE? 81

(see the dashed line in Figure 33). But to what is this force ap¬
plied? Are we justified in calling it a force when one of its com¬
ponents is applied to the earth and the other to an entirely differ¬
ent body-the string? Evidently, in the given case, the concept of a
centrifugal force has no physical meaning.

STUDENT A: In what cases does the centrifugal force exist?

TEACHER: In the case of a satellite in orbit, for instance, when
only two bodies interact the earth and the satellite. The cen¬
tripetal force is the force with which the earth attracts the satel¬
lite. The centrifugal force is the force with which the satellite
attracts the earth.

STUDENT B: You said that Newton’s third law was not valid
for the resultant of real forces I think that in this case it will be
invalid also for the components of a real force. Is that true?

TEACHER: Yes, quite true. In this connection I shall cite an
example which has nothing in common with rotary motion. A
ball lies on a floor and touches a wall which makes an obtuse
angle with the floor (Figure 34).

Let us resolve the weight of the ball into two components: per¬
pendicular to the wall and parallel to the floor. We shall deal
with these two components instead of the weight of the ball. If
Newton’s third law were applicable to separate components, we
could expect a reaction of the wall counter balancing the com¬
ponent of the weight perpendicular to it. Then, the component
of the weight parallel to the floor would remain unbalanced and
the ball would have to have a horizontal acceleration. Obviously,
this is physically absurd.

STUDENT A: So far you have discussed uniform motion in a
circle. How do you deal with a body moving nonuniformly
in a circle? For instance, a body slides down from the top of a
vertically held hoop. While it slides along the hoop it is moving
in a circle. This cannot, however, be uniform motion because the
velocity of the body increases. What do you, do in such cases?

Figure 34: A ball on the floor
touching a wall at an obtuse
angle with the floor.

82 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: If a body moves in a circle at uniform velocity, the
resultant of all forces applied to the body must be directed to
the centre; it imparts centripetal acceleration to the body. In the
more general case of nonuniform motion in a circle, the resultant
force is not directed strictly toward the centre. In this case, it
has a component along a radius toward the centre and another
component tangent to the trajectory of the body (i.e. to the
circle). The first component is responsible for the centripetal
acceleration of the body, and the second component, for the
so-called tangential acceleration, associated with the change
in velocity. It should be pointed out that since the velocity of
the body changes, the centripetal acceleration v 2 / r must also
change.

STUDENT A: Does that mean that for each instant of time the
centripetal acceleration will be determined by the formula a =
v 2 1 r , where v is the instantaneous velocity?

TEACHER: Exactly. While the centripetal acceleration is constant
in uniform motion in a circle, it varies in the process of motion
in nonuniform motion in a circle.

STUDENT A: What does one do to find out just how the velocity
v varies in nonuniform rotation?

TEACHER: Usually, the law of conservation of energy is resorted
to for this purpose. Let us consider a specific example. Assume
that a body slides without friction from the top of a vertically
held hoop of radius R. With what force will the body press on
the hoop as it passes a point located at a height h cm below the
top of the hoop? The initial velocity of the body at the top of the
hoop equals zero.

First of all, it is necessary to find what forces act on the body.

STUDENT A: Two forces act on the body: the weight P and the
bearing reaction N. They are shown in Figure 35.

TEACHER: Correct. What are you going to do next?

HOW DO YOU DEAL WITH MOTION IN A CIRCLE? 83

Figure 35: Resolution of
forces acting on a body in
nonuniform circular motion.

STUDENT A: I’m going to do as you said. I shall find the resultant
of these two forces and resolve it into two components: one
along the radius and the other tangent to the circle.

TEACHER: Quite right. Though it would evidently be simpler to
start by resolving the two forces applied to the body in the two
directions instead of finding the resultant, the more so because it
will be necessary to resolve only one force - the weight.

STUDENT A: My resolution of the forces is shown in Figure 35.

TEACHER: Force P 2 is responsible for the tangential acceleration
of the body, it does not interest us at present. The resultant of
forces P\ and N causes the centripetal acceleration of the body,

i.e.

P { -N =

mv 2

~R

(35)

The velocity of the body at the point we are interested in (point
A in Figure 35) can be found from the law of conservation of
energy

(36)

Combining equations (35) and (36) and taking into consideration
that

'R-h'

P { = P cos a = P

R

we obtain

j(R-h)-N

2 Ph
~i R ~

84 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

The sought-for force with which the body presses on the hoop is
equal, according to Newton’s third law, to the bearing reaction

(37)

STUDENT B: You assume that at point A the body is still on the
surface of the hoop. But it may fly off the hoop before it gets to
point A.

TEACHER: We can find the point at which the body leaves the
surface of the hoop. This point corresponds to the extreme case
when the force with which the body presses against the hoop
is reduced to zero. Consequently, in equation (37), we assume
N = 0 and solve for h, i.e, the vertical distance from the top of
the hoop to the point at which the body flies off. Thus

(38)

If in the problem as stated the value of b complies with the
condition h < h 0 then the result of equation (37) is correct; if,
instead, h > h 0 , then N = 0.

STUDENT A: As far as I can make out, two physical laws, equa¬
tions (35) and (36), were used to solve this problem.

TEACHER: Very good that you point this out. Quite true, two
laws are involved in the solution of this problem: Newton’s
second law of motion [see equation (35)] and the law of conser¬
vation of energy [see equation (36)]. Unfortunately, examinees
do not always clearly understand just which physical laws they
employ in solving some problem or other. This, I think, is an
essential point.

Take, for instance, the following example. An initial velocity
v 0 is imparted to a body so that it can travel from point A to
point C. Two alternative paths leading from A to C are offered
(Figure 36 (a), (b)). In both cases the body must reach the same
height H , but in different ways. Find the minimum initial veloc¬
ity v 0 for each case. Friction can be neglected.

HOW DO YOU DEAL WITH MOTION IN A CIRCLE? 85

STUDENT B: I think that the minimum initial velocity should be
the same in both cases, because there is no friction and the same
height is to be reached. This velocity can be calculated from the
law of conservation of energy

mgH =

mv g

-- from which v n

2 0

V2gH

TEACHER: Your answer is wrong. You overlooked the fact that
in the first case, the body passes the upper point of its trajectory
when it is in a state of rotational motion. This means that at the
top point B (Figure 36 (a)) it will have a velocity v 1 determined
from a dynamics equation similar to equation (35). Since the
problem involves the finding of a minimum, we should consider
the extreme case when the pressure of the body on its support at
point B is reduced to zero. Then only the weight will be acting
on the body and imparting to it the centripetal acceleration.
Thus

mg

mv\

~R~

2 mv\
H

(39)

Adding to the dynamics equation (39) the energy equation

mv q

2

mv\

- V mgH

2 5

(40)

we find that the minimum initial velocity equals, y 5gif/2. In
the second case, the body may pass the top point at a velocity in¬
finitely close to zero and so we can limit ourselves to the energy
equation. Then your answer is correct.

STUDENT B: Now I understand. If in the first case the body had
no velocity at point B, it would simply fall off its track.

TEACHER: If in the first case the body had the initial velocity
v Q = \j2gH as you suggested, it would never reach point B but
would fall away from the track somewhat earlier. I propose that
you find the height h of the point at which the body would fall
away from the track if its initial velocity was v 0 = *j2gH.

B

Figure 36: A body travels from
point A to point C via two
different paths. The problem
is to find the minimum initial
velocity in the two cases.

STUDENT A: Please let me try to do this problem.

86 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 37: A body moving on
a circular track. The problem
is to find the point at which
it would fall away from the
track if the initial velocity is
= \/ 2 S H -

TEACHER: Certainly.

STUDENT A: At the point the body drops off the track the
bearing reaction is evidently equal to zero. Therefore, only
the weight acts on the body at this point. We can resolve the
weight into two components, one along the radius (mg cos a)
and the other perpendicular to the radius (mg sin a) as shown in
Figure 37 (point A is the point at which the body falls away from
the track). The component along the radius imparts a centripetal
acceleration to the body, determined by the equation

mv^

mgcosa= - (41)

where v 2 is the velocity of the body at point A. To find it we can
make use of the energy equation

mv 2

-b mgh

2 6

mv q
2

(42)

Combining the dynamics (41) and energy (42) equations, taking
into consideration that cos a = (h —R)/R we obtain

mg(h —R)

mvn

2 mgh

from which

After substituting Vq

h = 2v\ 4- gH

6 g

2 gH the final result is

h

(43)

HOW DO YOU DEAL WITH MOTION IN A CIRCLE?

87

TEACHER: Entirely correct. Note that you can use equation (43)
to find the initial velocity v 0 for the body to loop the loop. For
this we take h = H in equation (43). Then

H H+s H

6 g

From this we directly obtain the result previously determined

STUDENT A: Condition (43) was obtained for a body falling
off its track. How can it be used for the case in which the body
loops the loop without falling away?

TEACHER: Falling away at the very top point of the loop actually
means that the body does not fall away but passes this point,
continuing its motion in a circle.

STUDENT B: One could say that the body falls away as if only for
a single instant.

TEACHER: Quite true. In conclusion I propose the following
problem (Figure 38 (a)). A body lies at the bottom of an in¬
clined plane with an angle of inclination a. This plane rotates at
uniform angular velocity co about a vertical axis. The distance
from the body to the axis of rotation of the plane equals R. Find
the minimum coefficient k 0 (I remind you that this coefficient
characterizes the maximum possible value of the force of static
friction) at which the body remains on the rotating inclined
plane without sliding off.

Fet us begin as always with the question: what forces are applied
to the body?

STUDENT A: Three forces are applied to the body: the weight P,
bearing reaction N, and the force of friction iy r .

TEACHER: Quite correct. It’s a good thing that you didn’t add
the centripetal force. Now what are you going to do next?

(a)

P

Figure 38: A body lies on a ro¬
tating inclined plane. Problem
is to find the maximum value
of static friction coefficient at
which body remains on the
plane without sliding off.

88 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT A: Next, I shall resolve the forces in the directions
along the plane and perpendicular to it as shown in Figure 38 (b).

TEACHER: I’ll take the liberty of interrupting you at this point. I
don’t like the way you have resolved the forces. Tell me, in what
direction is the body accelerated?

STUDENT A: The acceleration is directed horizontally. It is
centripetal acceleration.

TEACHER: Good. That is why you should resolve the forces
horizontally (i.e. along the acceleration) and vertically (i.e. per¬
pendicular to the acceleration). Remember what we discussed in
§ 6 .

STUDENT A: Now I understand. The resolution of the forces in
the horizontal and vertical directions is shown in Figure 38 (c).
The vertical components of the forces counterbalance one an¬
other, and the horizontal components impart acceleration to the
body. Thus

N cos a + iy r sin a = P 1

Fj- r cos a — A? sin a

mv 2 |

-R- )

Taking into consideration that

„ 2 '

R

Ff r = k n N, ( — ) = co'R and m = ( —

we can rewrite these equations in the form

N (cos a + k o sin a) = P
N(k 0 cos a — sin a) = Pco 2 Rg

STUDENT B: You have only two equations and three unknowns:
k Q , P and N.

TEACHER: That is no obstacle. We don’t have to find all three
unknowns, only the coefficient k 0 . The unknowns P and N can
be easily eliminated by dividing the first equation by the second.

HOW DO YOU DEAL WITH MOTION IN A CIRCLE?

89

STUDENT A: After dividing we obtain

cos a + k 0 sin a g
k 0 cos a — sin a co 2 R

which we solve for the required coefficient

k 0 =

co 2 R cos a + g sin a

g cos a

co 2 Ri

(44)

TEACHER: It is evident from equation (44) that the condition
cos a — co 2 R sin a^j > 0

should hold true. This condition can also be written in the form

aaa< 3 R <45)

If condition (45) is not complied with, no friction force is capable
of retaining the body on the rotating inclined plane.

Problems

15. What is the ratio of the forces with which an army tank bears down on the
middle of a convex and of a concave bridge? The radius of curvature of the
bridge is 40 m in both cases and the speed of the tank is 45 km/h.

16. A body slides without friction from the height H = 60 cm and then loops
the loop of radius R = 20 cm (Figure 39). Find the ratio of the forces with
which the body bears against the track at points A, B and C.

Figure 39: A body slides from
a given height and then loops
on a circular track. Problem
is to find ratio of forces which
the body bears against the
track at points A, B and C. See
Problem 16.

17. A body can rotate in a vertical plane at the end of a string of length R.
What horizontal velocity should be imparted to the body in the top posi¬
tion so that the tension of the string in the bottom position is ten times as
great as the weight of the body?

90

QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

18. Calculate the density of the substance of a spherical planet if a satellite
rotates about it with a period T in a circular orbit at a distance from the
surface of the planet equal to one half of its radius R. The gravitational
constant is denoted by G.

19. A body of mass m can slide without friction along a trough bent in the
form of a circular arc of radius R. At what height h will the body be at rest
if the trough rotates at a uniform angular velocity co (Figure 40) about a
vertical axis? What force F does the body exert on the trough?

Figure 40: A body slides with¬
out friction on a uniformly
rotating circular trough. Prob¬
lem is to find the force the
body exerts on the trough and
the height at which the body
will be at rest. See Problem 19.

20. A hoop of radius R is fixed vertically on the floor. A body slides without
friction from the top of the hoop (Figure 41). At what distance / from the
point where the hoop is fixed will the body fall?

Figure 41: A body slides
without friction form top
of the hoop. Problem is to
find distance / from the point
where the hoop is fixes and the
body will fall. See Problem 20.

§ 9 How Do You Explain The Weightlessness Of
Bodies?

TEACHER: How do you understand the expression: At the equa¬
tor of a planet, a body weightless than at the poles?

STUDENT B: I understand it as follows. The attraction of a body
by the earth is less at the equator than at the poles for two rea¬
sons. In the first place, the earth is somewhat flattened at the
poles and therefore the distance from the centre of the earth is
somewhat less to the poles than to the equator. In the second
place, the earth rotates about its axis as a result of which the force
of attraction at the equator is weakened due to the centrifugal
effect.

STUDENT A: Please make your last remark a little clearer.

STUDENT B: You must subtract the centrifugal force from the
force of attraction.

STUDENT A: I don’t agree with you. Firstly, the centrifugal
force is not applied to a body travelling in a circle. We already
discussed that in the preceding section (§ 8). Secondly, even if
such a force existed it could not prevent the force of attraction
from being exactly the same as if there was no rotation of the
earth. The force of attraction equals GmM/r 1 and, as such, does
not change just because other forces may act on the body.

TEACHER: As you can see, the question of the “weightness of
bodies” is not as simple as it seems at first glance. That’s why it

92 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

is among the questions that examinees quite frequently fail to an¬
swer correctly. As a matter of fact, if we agree on the definition

that the “weight of a body” is the force with which the body is
attracted by the earth, i.e. the force GmM /r 2 , then the reduc¬
tion of weight at the equator should be associated only with the

flattening at the poles (or bulging at the equator).

STUDENT B: But you cannot disregard rotation of the earth!

TEACHER: I fully agree with you. But first I wish to point out
that usually, in everyday life, the “weight of a body” is under¬
stood to be, not the force with which it is attracted to the earth,
and this is quite logical, but the force measured by a spring bal¬
ance, i.e. the force with which the body bears against the earth.
In other words, the bearing reaction is measured (the force with
which a body bears against a support is equal to the bearing
reaction according to Newton’s third law). It follows that the
expression “a body weighs less at the equator than at the poles”
means that at the equator it bears against its support with a lesser
force than at the poles.

Let us denote the force of attraction at the poles by P 1 and at the
equator by P 2 , the bearing reaction at the poles by and at the
equator by P 2 . At the poles the body is at rest, and at the equator
it travels in a circle. Thus

P, —N, =0

P 2 -N 2 = ma t

where a c p is the centripetal acceleration. We can rewrite these
equations in the form

(46)

Here it is clear that N 2 is less than N l since, firstly, P 2 is less than
Pi (from the effect of the flattening at the poles) and, secondly,
we subtract from P 2 the quantity ma c p (the effect of rotation of
the earth).

HOW DO YOU EXPLAIN THE WEIGHTLESSNESS OF BODIES? 93

STUDENT B: So, the expression “a body has lost half of its
weight” does not mean that the force with which it is attracted to
the earth (or any other planet) has been reduced by one half?

TEACHER: No, it doesn’t. The force of attraction may not change
at all. This expression means that the force with which the body
bears against its support (in other words, the bearing reaction)
has been reduced by one half.

STUDENT B: But then it follows that I can dispose of the “ weight -
ness” of a body quite freely. What can prevent me from digging
a deep pit under the body and letting it fall into the pit together
with its support? In this case, there will be no force whatsoever
bearing on the support. Does that mean that the body has com¬
pletely “lost its weight”? That it is in a state of weightlessness?

TEACHER: You have independently come to the correct conclu¬
sion. As a matter of fact, the state of weightlessness is a state of
fall of a body. In this connection I wish to make several remarks.

I have come across the interpretation of weightlessness as a state
in which the force of attraction of the earth is counterbalanced
by some other force. In the case of a satellite, the centrifugal
force was suggested as this counterbalancing force. The statement
was as follows: the force with which the satellite is attracted by
the earth and the centrifugal force counter balance each other
and, as a result, the resultant force applied to the satellite equals
zero, and this corresponds to weightlessness.

You now understand, of course, that such a treatment is incor¬
rect if only because no centrifugal force acts on the satellite.
Incidentally, if weightlessness is understood to be a state in which
the force of attraction is counterbalanced by some other force,
then it would be more logical to consider a body weightless
when it simply rests on a horizontal plane. This is precisely one
of the cases where the weight is counterbalanced by the bearing
reaction! Actually, no counter balancing of the force of attraction
is required for weightlessness. On the contrary, for a body to
become weightless, it is necessary to provide conditions in which
no other force acts on it except attraction. In other words, it is

94 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

necessary that the bearing reaction equal zero. The motion of a
body subject to the force of attraction is the falling of the body.
Consequently, weightlessness is a state of falling, for example the
falling of a lift in a mine shaft or the orbiting of the earth by a
satellite.

STUDENT A: In the preceding section (§ 8) you mentioned that
the orbiting of a satellite about the earth is none other than its
falling to the earth for an indefinitely long period of time.

TEACHER: That the motion of a satellite about the earth is falling
can be shown very graphically in the following way. Imagine
that you are at the top of a high mountain and throw a stone
horizontally. We shall neglect the air resistance. The greater the
initial velocity of the stone, the farther it will fall. Figure 42 (a)
illustrates how the trajectory of the stone gradually changes
with an increase in the initial velocity. At a certain velocity Vy
the trajectory of the falling stone becomes a circle and the stone
becomes a satellite of the earth. The velocity Vy is called the
circular orbital velocity. It is found from equation (34)

If the radius r of the satellite’s orbit is taken approximately equal
to the radius of the earth then Vy « 8 km/s.

STUDENT A: What will happen if in throwing a stone from the
mountain top we continue to increase the initial velocity?

TEACHER: The stone will orbit the earth in a more and more
elongated ellipse (Figure 42 (b)). At a certain velocity v 2 the
trajectory of the stone becomes a parabola and the stone ceases
to be a satellite of the earth. The velocity v 2 is called the escape
velocity. According to calculations, v 2 « 11 km/s. This is about
\fl times v x .

STUDENT A: You have defined the state of weightlessness as a
state of fall. However, if the initial velocity of the stone reaches
the escape velocity, the stone will leave the earth. In this case,

Figure 42: Trajectory of a body
around Earth with different
initial velocities.

HOW DO YOU EXPLAIN THE WEIGHTLESSNESS OF BODIES? 95

you cannot say that it is falling to the earth. How, then, can you
interpret the weightlessness of the stone?

TEACHER: Very simply. Weightlessness in this case is the falling
of the stone to the sun.

STUDENT A: Then the weightlessness of a spaceship located
somewhere in interstellar space is to be associated with the falling
of the ship in the gravitational field of some celestial body?

TEACHER: Exactly.

STUDENT B: Still, it seems to me that the definition of weightless¬
ness as a state of falling requires some refinement. A parachutist
also falls, but he has none of the sensations associated with
weightlessness.

TEACHER: You are right. Weightlessness is not just any kind of
falling. Weightlessness is the so-called free fall, i.e. the motion
of a body subject only (!) to the force of gravity. I have already
mentioned that for a body to become weightless it is necessary
to create conditions under which no other force, except the
force of attraction, acts on the body. In the case of the fall of a
parachutist, there is an additional force, the resistance of the air.

Problems

21. Calculate the density of the substance of a spherical planet where the daily
period of rotation equals 10 hours, if it is known that bodies are weightless
at the equator of the planet.

The role of the physical laws of conservation can scarcely be overesti¬
mated. They constitute the most general rules established by mankind
on the basis of the experience of many generations. Skillful application
of the laws of conservation enables many problems to be solved with
comparative ease. Let us consider examples concerning the laws of
conservation of energy and momentum.

§ 10 Can You Apply The Laws Of Conservation
Of Energy A nd Linear Momentum?

TEACHER: To begin with I wish to propose several simple prob¬
lems.

The first problem: Bodies slide without friction down two in¬
clined planes of equal height H but with two different
angles of inclination a 1 and a 2 - The initial velocity of the
bodies equals zero. Find the velocities of the bodies at the
end of their paths.

The second problem: We know that the formula expressing the
final velocity of a body in terms of the acceleration and
distance travelled v = filas refers to the case when there
is no initial velocity. What will this formula be if the body
has an initial velocity vfil

The third problem: A body is thrown from a height H with a

horizontal velocity of v 0 . Find its velocity when it reaches
the ground.

The fourth problem: A body is thrown upward at an angle a
to the horizontal with an initial velocity v 0 . Find the
maximum height reached in its flight.

STUDENT A: I shall solve the first problem in the following way.
We first consider one of the inclined planes, for instance the one
with the angle of inclination aq. Two forces are applied to the
body: the force of gravity P and the bearing reaction Aj. We

100 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

resolve the force P into two components, one along the plane
( P sin a j) and the other perpendicular to it ( P cos a{). We then
write the equations for the forces perpendicular to the inclined
plane

Pcosa l —N l = 0
and for the forces along the plane

p • Pa i

Psma 1 =-

g

where a l is the acceleration of the body. From the second equa¬
tion we find that a l = g sin a v The distance travelled by the
body is H/ sinoq. Next, using the formula mentioned in the
second problem, we find that the velocity at the end of the path
is

- /2g77sina,

v 1 = x /a 1 s 1 = a - : -= y/2gH

V sinaj

Since the final result does not depend upon the angle of inclina¬
tion, it is also applicable to the second plane inclined at the angle
a 2-

To solve the second problem I shall make use of the well known
kinematic relationships

v = v 0 + at

at 2

s = v 0 t H-

0 2

From the first equation we find that t = (v — v 0 )/a. Substituting
this for t in the second equation we obtain

v 0 {v- v 0 ) a (v-v 0 ) 2

S = - + a-2-

a 2 a z

or

2 sa = 2 vqV — 2v£ +v 2 — 2v§v + Vq

v = J las +

( 48 )

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM?

101

To solve the third problem, I shall first find the horizontal
and vertical v 2 components of the final velocity. Since the body
travels at uniform velocity in the horizontal direction, v 1 = v 0 .

In the vertical direction the body travels with acceleration g
but has no initial velocity. Therefore, we can use the formula
v 2 = \j2gH. Since the sum of the squares of the sides of a right
triangle equals, the square of the hypotenuse, the final answer is

v = yj v\ + v\ = yj v\ + 2 gH (49)

The fourth problem has already been discussed in § 5. It is nec¬
essary to resolve the initial velocity into the horizontal (v 0 cos a)
and vertical (t^ sin or) components. Then we consider the vertical
motion of the body and, first of all, we find the time fj of ascent
from the formula for the dependence of the velocity on time in
uniformly decelerated motion (y v = v 0 s'ma — gt), taking into
account that at t = t x the vertical velocity of the body vanishes.
Thus v 0 s'ma — gfj = 0, from which 1 1 = (f 0 /g)sina. The time
tj being known, we find the height H reached from the formula
of the dependence of the distance travelled on time in uniformly
decelerated motion. Thus

TT t *0 ■

H = v 0 f i sm a -= — sin a

2 2 g

TEACHER: In all four cases you obtained the correct answers. I
am not, however, pleased with the way you solved these prob¬
lems. They could all have been solved much simpler if you had
used the law of conservation of energy. You can see for yourself.

First problem.

The law of conservation of energy is of the form mgH =
mv 1 /2 (the potential energy of the body at the top of the
inclined plane is equal to its kinetic energy at the bottom).
From this equation we readily find the velocity of the
body at the bottom

v = \/2gH

102 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Second problem.

In this case, the law of conservation of energy is of the
form mv^/2 + mas = mv 2 / 2, where mas is the work
done by the forces in imparting the acceleration a to the
body. This leads immediately to v\ + 2a s = v 2 or, finally,
to

v = y J las + V 2

Third problem.

We write the law of conservation of energy as mgH +
mv q2 = mv 2 /2. Then the result is

v = yj 2 gH + v 2

Fourth problem.

At the point the body is thrown its energy equals mv^/2.
At the top point of its trajectory, the energy of the body
is mgH + mv 2 /2. Since the velocity v 1 at the top point
equals v 0 cos a, then, using the law of conservation of
energy

mv/ mv/ 7

-= mgH H-- cos a

2 2

we find that

"-(^)(‘- cos2 “)

or, finally

STUDENT A: Yes, it’s quite clear to me that these problems can
be solved in a much simpler way. It didn’t occur to me to use the
law of conservation of energy.

TEACHER: Unfortunately, examinees frequently forget about
this law. As a result they begin to solve such problems by more
cumbersome methods, thus increasing the probability of errors.
My advice is: make more resourceful and extensive use of the
law of conservation of energy. This poses the question: how
skillfully can you employ this law?

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM?

103

STUDENT A: It seems to me that no special skill is required; the
law of conservation of energy as such is quite simple.

TEACHER: The ability to apply a physical law correctly is not de¬
termined by its complexity or simplicity. Consider an example.
Assume that a body travels at uniform velocity in a circle in a
horizontal plane. No friction forces operate. The body is subject
to a centripetal force. What is the work done by this force in one
revolution of the body?

STUDENT A: Work is equal to the product of the force by the
distance through which it acts. Thus, in our case, it equals
( mv 2 /R)2 tcR = Inmv 2 , where R is the radius of the circle
and m and v are the mass and velocity of the body.

TEACHER: According to the law of conservation of energy; work
cannot completely disappear. What has become of the work you
calculated?

STUDENT A: It has been used to rotate the body.

TEACHER: I don’t understand. State it more precisely.

STUDENT A: It keeps the body on the circle.

TEACHER: Your reasoning is wrong. No work at all is required to
keep the body on the circle.

STUDENT A: Then I don’t know how to answer your question.

TEACHER: Energy imparted to a body can be distributed, as
physicists say, among the following “channels”: (1) increasing the
kinetic energy of the body; (2) increasing its potential energy;

(3) work performed by the given body on other bodies; and

(4) heat evolved due to friction. Such is the general principle
which not all examinees understand with sufficient clarity. Now
consider the work of the centripetal force. The body travels

at a constant velocity and therefore its kinetic energy is not
increased. Thus the first channel is closed. The body travels in
a horizontal plane and, consequently, its potential energy is not
changed. The second channel is also closed. The given body does

104 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

not perform any work on other bodies, so that the third channel
is closed. Finally, all kinds of friction have been excluded. This
closes the fourth and last channel.

STUDENT A: But then there is simply no room for the work of
the centripetal force, or is there?

TEACHER: As you see, none. It remains now for you to declare
your position on the matter. Either you admit that the law your
this your of conservation of energy is not valid, and then all
troubles are gone, or you proceed from the validity of law and
then_However, try to find the way out of difficulties.

STUDENT A: I think that it remains to conclude that the cen¬
tripetal force performs no work whatsoever.

TEACHER: That is quite a logical conclusion. I want to point out
that it is the direct consequence of the law of conservation of
energy.

STUDENT B: All this is very well, but what do we do about the
formula for the work done by a body?

TEACHER: In addition to the force and the distance through
which it acts, this formula should also contain the cosine of
the angle between the direction of the force and the velocity
A = Fs cos a. In the given case, cos a = 0.

STUDENT A: Oh yes, I entirely forgot about this cosine.

TEACHER: I want to propose another example. Consider com¬
municating vessels connected by a narrow tube with a stopcock.
Assume that at first all the liquid is in the left vessel and its height
is H (Figure 43(a)). Then we open the stopcock and the liquid
flows from the left into the right vessel. The flow ceases when
there is an equal level of Hj 2 in each vessel (Figure 43(b)).

Let us calculate the potential energy of the liquid in the initial
and final positions. For this we multiply the weight of the liquid
in each vessel by one half of the column of liquid. In the initial

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM? 105

position the potential energy equalled PH/ 2, and in the final one
it is

Thus in the final state, the potential energy of the liquid turns
out to be only one half of that in the initial state. Where has one
half of the energy disappeared to?

STUDENT A: I shall attempt to reason as you advised. The poten-
PH

tial energy —j— could be used up in performing work on other
bodies, on heat evolved in friction and on the kinetic energy of
the liquid itself. Is that right?

TEACHER: Quite correct. Continue.

STUDENT A: In our case, the liquid flowing from one vessel
to the other does not perform any work on other bodies. The
liquid has no kinetic energy in the final state because it is in a
state of rest. Then, it remains to conclude that one half of the
potential energy has been converted into heat evolved in friction.
True, I don’t have a very clear idea of what kind of friction it is.

TEACHER: You reasoned correctly and came to the right con¬
clusion. I want to add a few words on the nature of friction.

One can imagine that the liquid is divided into layers, each char¬
acterizing a definite rate of flow. The closer the layer to the
walls of the tube, the lower its velocity. There is an exchange
of molecules between the layers, as a result of which molecules
with a higher velocity of ordered motion find themselves among
molecules with a lower velocity of ordered motion, and vice
versa. As a result, the “faster” layer has an accelerating effect
on the “slower” layer and, conversely, the “slower” layer has a
decelerating effect on the “faster” layer.

This picture allows us to speak of the existence of a peculiar
internal friction between the layers. Its effect is stronger with
a greater difference in the velocities of the layers in the middle
part of the tube and near the walls. Note that the velocity of
the layers near the walls is influenced by the kind of interaction

(a)

* IT

(b)

(c)

— 3 :

Figure 43: Problem with
potential energy of a liquid
column in a communicating
vessel.

106 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

between the molecules of the liquid and those of the walls. If the
liquid wets the tube then the layer directly adjacent to the wall is
actually stationary.

STUDENT A: Does this mean that in the final state the tempera¬
ture of the liquid should be somewhat higher than in the initial
state?

TEACHER: Yes, exactly so. Now we shall change the conditions
of the problem to some extent. Assume that there is no interac¬
tion between the liquid and the tube walls. Hence, all the layers
will flow at the same velocity and there will be no internal fric¬
tion. How then will the liquid flow from one vessel to the other?

STUDENT A: Here the potential energy will be reduced owing
to the kinetic energy acquired by the liquid. In other words,
the state illustrated in Figure 43(b) is not one of rest. The liquid
will continue to flow from the left vessel to the right one until it
reaches the state shown in Figure 43(c). In this state the potential
energy is again the same as in the initial state Figure 43(a).

TEACHER: What will happen to the liquid after this?

STUDENT A: The liquid will begin to flow back in the reverse
direction, from the right vessel to the left one. As a result, the
levels of the liquid will fluctuate in the communicating vessels.

TEACHER: Such fluctuations can be observed, for instance, in
communicating glass vessels containing mercury. We know that
mercury does not wet glass. Of course these fluctuations will be
damped in the course of time, since it is impossible to completely
exclude the interaction between the molecules of the liquid and
those of the tube walls.

STUDENT A: I see that the law of conservation of energy can be
applied quite actively.

TEACHER: Here is another problem for you. A bullet of mass m,
travelling horizontally with a velocity v 0 , hits a wooden block
of mass M, suspended on a string, and sticks in the block. To

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM? 107

what height H will the block rise, after the bullet hits it, due to
deviation of the string from the equilibrium position (Figure 44)?

Figure 44: A bullet hits a block
of wood and sticks in it. The
problem is to find the height to
which the block will rise after
the impact.

STUDENT A: We shall denote by v 1 the velocity of the block
with the bullet immediately after the bullet hits the block. To
find this velocity we make use of the law of conservation of

energy. Thus

2 „,2
mv„ , „ v.

0 =(m+M 1

2 2

(50)

from which

/ m

Vl ~ V °\ m+M

(51)

This velocity being known, we find the sought-for height H by
again resorting to the law of conservation of energy

(m + M) gH = {m + M)— (52)

Equations (50) and (52) can be combined

mvi

(m +M)gH = -^

from which

H =

„,2

m v Q

(m +M) 2g

(53)

TEACHER: (to STUDENT B) What do you think of this solution?

STUDENT B: I don’t agree with it. We were told previously that
in such cases the law of conservation of momentum is to be used.

108 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Therefore, instead of equation (50) I would have used a different
relationship

mv 0 = (m+M)v 1 (54)

(the momentum of the bullet before it hits the block is equal to
the momentum of the bullet and block afterward). From this it
follows that

(m +M)

(55)

If we now use the law of conservation of energy (52) and substi¬
tute the result of equation (55) into (52) we obtain

H =

m

m +M ) 2 g

2v l

(56)

TEACHER: We have two different opinions and two results. The
point is that in one case the law of conservation of kinetic energy
is applied when the bullet strikes the block, and in the other case,
the law of conservation of momentum. Which is correct? (to
STUDENT A): What can you say to justify your position?

STUDENT A: It didn’t occur to me to use the law of conservation
of momentum.

TEACHER: (to STUDENT B): And what do you say?

STUDENT B: I don’t know how to substantiate my position. I
remember that in dealing with collisions, the law of conservation
of momentum is always valid, while the law of conservation of
energy does not always hold good. Since in the given case these
laws lead to different results, my solution is evidently correct.

TEACHER: As a matter of fact, it is indeed quite correct. It is,
however, necessary to get a better insight into the matter. A
collision after which the colliding bodies travel stuck together
(or one inside the other) is said to be a “completely inelastic
collision”. Typical of such impacts is the presence of permanent
set in the colliding bodies, as a result of which a certain amount
of heat is evolved. Therefore, equation (50), referring only to
the kinetic energy of bodies, is inapplicable. In our case, it is

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM?

109

necessary to employ the law of conservation of momentum (54)
to find the velocity of the box with the bullet after the impact.

STUDENT A: Do you mean to say that the law of conservation of
energy is not valid for a completely inelastic collision? But this
law is universal.

TEACHER: There is no question but that the law of conserva¬
tion of energy is valid for a completely inelastic collision as well.
The kinetic energy is not conserved after such a collision. I
specifically mean the kinetic energy and not the whole energy.
Denoting the heat evolved in collision by Q, we can write the
following system of laws of conservation referring to the com¬
pletely inelastic collision discussed above

mv Q
mv g

~Y~

(m -I -M)v l
(m + M)v ?

-—^—— l +q

(57)

Here the first equation is the law of conservation of momentum,
and the second is the law of conservation of energy (including
not only mechanical energy, but heat as well). The system of
equations (57) contains two unknowns: v 1 and Q. After de¬
termining v 1 from the first equation, we can use the second
equation to find the evolved heat Q

mv£ (m + M)m 2 v^ mv g / m

2 2 (m+M) 2 2 V m+M

It is evident from this equation that the larger the mass M , the
more energy is converted into heat. In the limit, for an infinitely
large mass. M , we obtain Q = mv q/ 2, i.e. all the kinetic energy
is converted into heat. This is quite natural: assume that the
bullet sticks in a wall.

STUDENT A: Can there be an impact in which no heat is evolved?

TEACHER: Yes, such collisions are possible. They are said to be
“perfectly elastic”. For instance, the impact of two steel balls can
be regarded as perfectly elastic with a fair degree of approxima¬
tion. Purely elastic deformation of the balls occurs and no heat

110 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

is evolved. After the collision, the balls return to their original
shape.

STUDENT A: You mean that in a perfectly elastic collision, the
law of conservation of energy becomes the law of conservation
of kinetic energy?

TEACHER: Yes, of course.

STUDENT A: But in this case I cannot understand how you can
reconcile the laws of conservation of momentum and of energy.
We obtain two entirely different equations for the velocity after
impact. Or, maybe, the law of conservation of momentum is not
valid for a perfectly elastic collision.

TEACHER: Both conservation laws are valid for a perfectly elastic
impact: for momentum and for kinetic energy. You have no rea¬
son to worry about the reconciliation of these laws because after
a perfectly elastic impact, the bodies fly apart at different veloc¬
ities. Whereas after a completely inelastic impact the colliding
bodies travel at the same velocity (since they stick together), after
an elastic impact each body travels at its own definite velocity.
Two unknowns require two equations.

Let us consider an example. Assume that a body of mass m
travelling at a velocity v 0 elastically collides with a body of
mass M at rest. Further assume that as a result of the impact
the incident body bounces back. We shall denote the velocity
of body m after the collision by v 1 and that of body M by v 2 -
Then the laws of conservation of momentum and energy can be
written in the form

mv 0 = Mv 2 — mv j
mv g mvI; mv\

~T~ ~ ~1T + ^T

>

(59)

Note the minus sign in the first equation. It is due to our assump¬
tion that the incident body bounces back.

STUDENT B: But you cannot always know beforehand in which
direction a body will travel after the impact. Is it impossible for

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM?

Ill

the body m to continue travelling in the same direction but at a
lower velocity after the collision?

TEACHER: That is quite possible. In such a case, we shall obtain a
negative velocity v 1 when solving the system of equations (59).

STUDENT B: I think that the direction of travel of body m after
the collision is determined by the ratio of the masses m and M.

TEACHER: Exactly. If m < M, body m will bounce back; at
m = M, it will be at rest after the collision; and at m > M , it will
continue its travel in the same direction but at a lower velocity.

In the general case, however, you need not worry about the
direction of travel. It will be sufficient to assume some direction
and begin the calculations. The sign of the answer will indicate
your mistake, if any.

STUDENT B: We know that upon collision the balls may fly apart
at an angle to each other. We assumed that motion takes place
along a single straight line. Evidently, this must have been a
special case.

TEACHER: You’ are right. We considered what is called a central
collision in which the balls travel before and after the impact
along a line passing through their centres. The more general case
of the off-centre collision will be dealt with later. For the time
being, I’d like to know if everything is quite clear.

STUDENT A: I think I understand now. As I see it, in any colli¬
sion (elastic or inelastic), two laws of conservation are applicable:
of momentum and of energy. Simply the different nature of the
impacts leads to different equations for describing the conserva¬
tion laws. In dealing with inelastic collisions, it is necessary to
take into account the heat evolved on impact.

TEACHER: Your remarks are true and to the point.

STUDENT B: So far as I understand it, completely elastic and
perfectly inelastic collisions are the two extreme cases. Are they
always suitable for describing real cases?

112 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: You are right in bringing up this matter. The cases
of collision we have considered are extreme ones. In real colli¬
sions some amount of heat is always generated (no ideally elastic
deformation exists) and the colliding bodies may fly apart with
different velocities. In many cases, however, real collisions are
described quite well by means of simplified models: completely
elastic and perfectly inelastic collisions.

Now let us consider an example of an off-centre elastic collision.
A body in the form of an inclined plane with a 45° angle of
inclination lies on a horizontal plane. A ball of mass m, flying
horizontally with a velocity v 0 , collides with the body (inclined
plane), which has a mass of M. As a result of the impact, the
ball bounces vertically upward and the body M begins to slide
without friction along the horizontal plane. Find the velocity
with which the ball begins its vertical travel after the collision.
(Figure 45).

Which of you wishes to try your hand at this problem?

Figure 45: A ball collides with
another body at rest in the
form of an inclined plane. The
problem is to find the velocity
of the ball after the impact.

STUDENT B: Allow me to. Let us denote the sought-for velocity
of the ball by v 1 and that of body M by v 2 - Since the collision
is elastic, I have the right to assume that the kinetic energy is
conserved. Thus

mv q mv j m-Vj

(60)

I need one more equation, for which I should evidently use the
law of conservation of momentum. I shall write it in the form

mv 0 =Mv 2 + mv 1 (61)

True, Fm not so sure about this last equation because velocity v 1
is at right angles to velocity v 2 .

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM? 113

TEACHER: Equation (60) is correct. Equation (61) is incorrect,
just as you thought. You should remember that the law of con¬
servation of momentum is a vector equation, since the momen¬
tum is a vector quantity having the same direction as the velocity.

True enough, when all the velocities are directed along a single
straight line, the vector equation can be replaced by a scalar one.

That is precisely what happened when we discussed central col¬
lisions. In the general case, however, it is necessary to resolve all
velocities in mutually perpendicular directions and to write the
law of conservation of momentum for each of these directions
separately (if the problem is considered in a plane, the vector
equation can be replaced by two scalar equations for the pro¬
jections of the momentum in the two mutually perpendicular
directions).

For the given problem we can choose the horizontal and vertical
directions. For the horizontal direction, the law of conservation
of momentum is of the form

(62)

mv o = Mv 2

From equations (60) and (62) we find the velocity

M — m

m

STUDENT B: What do we do about the vertical direction?

TEACHER: At first sight, it would seem that the law of conserva¬
tion of momentum is not valid for the vertical direction. Actu¬
ally it is. Before the impact there were no vertical velocities; after
the impact, there is a momentum mv lt directed vertically up¬
wards. We can readily see that still another body participates in
the problem: the earth. If it was not for the earth, body M would
not travel horizontally after the collision. Let us denote the mass
of the earth by M ^ and the velocity it acquires as a result of the
impact by v 0 .

The absence of friction enables us to treat the interaction be¬
tween the body M and the earth as taking place only in the

114 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

vertical direction. In other words, the velocity t;® of the earth
is directed vertically downwards. Thus, the participation of the
earth in our problem doesn’t change the form of equation (62),
but leads to an equation which describes the law of conservation
of momentum for the vertical direction

mv 1 — T/®f® = 0

(63)

STUDENT B: Since the earth also participates in this problem it
will evidently be necessary to correct the energy relation (60).

TEACHER: Just what do you propose to do to equation (60)?

STUDENT B: I wish to add a term concerning the motion of the
earth after the impact

mvX

mvt

Mvj

M a

(64)

TEACHER: Your intention is quite logical. There is, however, no
need to change equation (60). As a matter of fact, it follows from
equation (63) that the velocity of the earth is

mv j

Since the mass T/® is practically infinitely large, the velocity
f® of the earth after the impact is practically equal to zero.
Now, let us rewrite the term Tf® v^/2 in equation (64) to obtain
the form (T/®f®)f®/ 2 . The quantity M ®f® in this product
has, according to equation (63), a finite value. If this value is
multiplied by zero (in the given case by t>®) the product is also
zero.

From this we can conclude that the earth participates very pe¬
culiarly in this problem. It acquires a certain momentum, but
at the same time, receives practically no energy. In other words,
it participates in the law of conservation of momentum, but
does not participate in the law of conservation of energy. This
circumstance is especially striking evidence of the fact that the
laws of conservation of energy and of momentum are essentially
different, mutually independent laws.

CAN YOU APPLY THE LAWS OF CONSERVATION OF ENERGY AND LINEAR MOMENTUM?

115

Problems

22. A body with a mass of 3 kg falls from a certain height with an initial ve¬
locity of 2 m/s, directed vertically downward. Find the work done to
overcome the forces of resistance during 10 s if it is known that the body
acquired a velocity of 50 m/s at the end of the 10 s interval. Assume that the
force of resistance is constant.

23. A body slides first down an inclined plane at an angle of 30° and then along
a horizontal surface. Determine the coefficient of friction if it is known
that the body slides along the horizontal surface the same distance as along
the inclined plane.

24. Calculate the efficiency of an inclined plane for the case when a body slides
off it at uniform velocity.

25. A ball of mass m and volume V drops into water from a height H, plunges
to a depth h and then jumps out of the water (the density of the ball is

less than that of water). Find the resistance of the water (assuming it to be
constant) and the height hy to which the ball ascends after jumping out of
the water. Neglect air resistance. The density of water is denoted by p w .

26. A railway car with a mass of 50 tons, travelling with a velocity of 12 km/h,
runs into a flatcar with a mass of 30 tons standing on the same track. Find
the velocity of joint travel of the railway car and flatcar directly after the
automatic coupling device operates. Calculate the distance travelled by the
two cars after being coupled if the force of resistance is 5 per cent of the
weight.

27. A cannon of mass M, located at the base of an inclined plane, shoots a shell
of mass m in a horizontal direction with an initial velocity v Q . To what
height does the cannon ascend the inclined plane as a result of recoil if the
angle of inclination of the plane is a and the coefficient of friction between
the cannon and the plane is k}

28. Two balls of masses M and 2 M are hanging on threads of length / fixed at
the same point. The ball of mass M is pulled to one side through an angle
of a and is released after imparting to it a tangential velocity of v Q in the
direction of the equilibrium position. To what height will the balls rise
after collision if: (a) the impact is perfectly elastic, and (b) if it is completely
inelastic (the balls stick together as a result of the impact)?

29. A ball of mass M hangs on a string of length /. A bullet of mass m, flying
horizontally, hits the ball and sticks in it. At what minimum velocity must
the bullet travel so that the ball will make one complete revolution in a
vertical plane?

116 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

30. Two wedges with angles of inclination equal to 45° and each of mass M
lie on a horizontal plane (Figure 46). A ball of mass m (m « M) drops
freely from the height H. It first strikes one wedge and then the other and
bounces vertically upward. Find the height to which the ball bounces.
Assume that both impacts are elastic and that there is no friction between
the wedges and the plane.

31. A wedge with an angle of 30° and a mass M lies on a horizontal plane. A
ball of mass m drops freely from the height H, strikes the wedge elastically
and bounces away at an angle of 30° to the horizontal. To what height does
the ball ascend? Neglect friction between the wedge and the horizontal
plane.

Figure 46: A ball strikes one
of the two edges after being
dropped from a height. First it
strikes the second wedge and
then goes vertically up. The
problem is to find the height to
which the ball rises.

The world about us is full of vibrations and waves. Remember this when
you study the branch of physical science devoted to these phenom¬
ena. Let us discuss harmonic vibrations and, as a special case, the vi¬
brations of a mathematical pendulum. We shall analyse the behaviour
of the pendulum in non-inertial frames of reference.

§ 11 Can You Deal With Harmonic Vibrations?

TEACHER: Some examinees do not have a sufficiently clear un¬
derstanding of harmonic vibrations. First let us discuss their
definition.

STUDENT A: Vibrations are said to be harmonic if they obey the
sine law: the deviation x of a body from its equilibrium position
varies with time as follows

x = A sin(to t + a) (65)

where A is the amplitude of vibration (maximum deviation of
the body from the position of equilibrium), to is the circular
frequency (to = 2nF where T is the period of vibration), and a
is the initial phase (it indicates the deviation of the body from the
position of equilibrium at the instant of time t = 0). The idea of
harmonic vibrations is conveyed by the motion of the projection
of a point which rotates at uniform angular velocity to in a circle
of radius A (Figure 47).

STUDENT B: I prefer another definition of harmonic vibrations.
As is known, vibrations occur due to action of the restoring
force, i.e. a force directed toward the position of equilibrium and
increasing as the body recedes from the equilibrium position.
Flarmonic vibrations are those in which the restoring force F is
proportional to the deviation x of the body from the equilibrium
position. Thus

F = kx (66)

Such a force is said to be “elastic”.

Figure 47: Simple harmonic
motion and its relations.

120 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: I am fully satisfied with both proposed definitions.

In the first case, harmonic vibrations are defined on the basis of
how they occur; in the second case, on the basis of their cause.

In other words, the first definition uses the space-time (kine¬
matic) description of the vibrations, and the second, the causal
(dynamic) description.

STUDENT B: But which of the two definitions is preferable? Or,
maybe, they are equivalent?

TEACHER: No, they are not equivalent, and the first (kinematic)
is preferable. It is more complete.

STUDENT B: But whatever the nature of the restoring force, it
will evidently determine the nature of the vibrations. I don’t
understand, then, why my definition is less complete.

TEACHER: This is not quite so. The nature of the restoring force
does not fully determine the nature of the vibrations.

STUDENT A: Apparently, now is the time to recall that the nature
of the motion of a body at a given instant is determined not only
by the forces acting on the body at the given instant, but by the
initial conditions as well, i.e. the position and velocity of the
body at the initial instant. We discussed this in § 4.

TEACHER: Absolutely correct. With reference to the case being
considered this statement means that the nature of the vibrations
is determined not only by the restoring force, but by the con¬
ditions under which these vibrations started. It is evident that
vibrations can be effected in various ways. For example, a body
can be deflected a certain distance from its equilibrium position
and then smoothly released. It will begin to vibrate. If the begin¬
ning of vibration is taken as the zero instant, then from equation
(65), we obtain a = tt/ 2, and the distance the body is deflected is
the amplitude of vibration. The body can be deflected different
distances from the equilibrium position, thereby setting different
amplitudes of vibration.

Another method of starting vibrations is to impart a certain ini¬
tial velocity (by pushing) to a body in a state of equilibrium. The

CAN YOU DEAL WITH HARMONIC VIBRATIONS?

121

body will begin to vibrate. Taking the beginning of vibration
as the zero point, we obtain from equation (65) that a = 0. The
amplitude of these vibrations depends upon the initial velocity
imparted to the body. It is evidently possible to propose innu¬
merable other, intermediate methods of exciting vibrations. For
instance, a body is deflected from its position of equilibrium
and, at the same time, is pushed or plucked, etc. Each of these
methods will set definite values of the amplitude A and the initial
phase a of the vibration.

STUDENT B: Do you mean that the quantities A and a do not
depend upon the nature of the restoring force?

TEACHER: Exactly. You manipulate these two quantities at your
own discretion when you excite vibrations by one or another
method. The restoring force, i.e. coefficient k in equation (66),
determines only the circular frequency co or, in other words, the
period of vibration of the body. It can be said that the period
of vibration is an intrinsic characteristic of the vibrating body,
while the amplitude A and the initial phase a depend upon the
external conditions that excite the given vibration.

Returning to the definitions of harmonic vibrations, we see that
the dynamic definition contains no information on either the
amplitude or initial phase. The kinematic definition, on the
contrary, contains information on these quantities.

STUDENT B: But if we have such a free hand in dealing with the
amplitude, maybe it is not so important a characteristic of a
vibrating body?

TEACHER: You are mistaken. The amplitude is a very important
characteristic of a vibrating body. To prove this, let us consider
an example. A ball of mass m is attached to two elastic springs
and accomplishes harmonic vibrations of amplitude A in the
horizontal direction Figure 48. The restoring force is determined
by the coefficient of elasticity k which characterizes the elastic
properties of the springs. Find the energy of the vibrating ball.

STUDENT A: To find the energy of the ball, we can consider its

122 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 48: Simple harmonic
motion and its relations.

position of extreme deflection (x = A). In this position, the
velocity of the ball equals zero and therefore its total energy is its
potential energy. The latter can be determined as the work done
against the restoring force F in displacing the ball the distance A
from its equilibrium position. Thus

W = FA

(67)

Next, taking into account that F = kA, according to equation
(66), we obtain

W = kA 2

TEACHER: You reasoned along the proper lines, but committed
an error. Equation (67) is applicable only on condition that
the force is constant. In the given case, force F varies with the
distance, as shown in Figure 49. The work done by this force

over the distance x = A is equal to the hatched area under the

( 68 )

2

Note that the total energy of a vibrating body is proportional to

F

Figure 49: Force varying with
distance.

CAN YOU DEAL WITH HARMONIC VIBRATIONS?

123

the square of the amplitude of vibration. This demonstrates what
an important characteristic of a vibrating body the amplitude
is. If 0 < x < A, then the total energy W is the sum of two
components-the kinetic and potential energies

mv 2 kx 2

~Y~ —

(69)

Equation (69) enables the velocity v of the vibrating ball to be
found at any distance x from the equilibrium position. My next
question is: what is the period of vibration of the ball shown in
Figure 48?

STUDENT B: To establish the formula for the period of vibration
it will be necessary to employ differential calculus.

TEACHER: Strictly speaking, you are right. However, if we si¬
multaneously use the kinematic and dynamic definitions of har¬
monic vibrations we can manage without differential calculus.

As a matter of fact, we can’ conclude from Figure 47, which is a
graphical expression of the kinematic definition, that the velocity
of the body at the instant it passes the equilibrium position is

v x = coA = (70)

Using the result of equation (68), following from the dynamic
definition, we can conclude that velocity tq can be found from
the energy relation

mv 2 _ kA 2

~Y~ ~ ~T~

(71)

(at the instant the ball passes the equilibrium position the entire
energy of the ball is kinetic energy). Combining equations (70)
and (71), we obtain

4n 2 A 2 m

T-

kA 2 ,

from which

T = 2nJ^

(72)

As mentioned previously, the period of vibration is determined
fully by the properties of the vibrating system itself, and is inde¬
pendent of the way the vibrations are set up.

124 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT A: When speaking of vibrations we usually deal, not
with a ball attached to springs, but with a pendulum. Can the
obtained results be generalized to include the pendulum?

TEACHER: For such generalization we must first find out what,
in the case of the pendulum, plays the role of the coefficient of
elasticity k. It is evident that a pendulum vibrates not due to an
elastic force, but to the force of gravity. Let us consider a ball
(called a bob in a pendulum) suspended on a string of length /.
We pull the bob to one side of the equilibrium position so that
the string makes an angle a, (Figure 50) with the vertical. Two
forces act on the bob: the force of gravity mg and the tension T
of the string. Their resultant is the restoring force. As is evident
from the figure, it equals mg sin a.

Figure 50: Analysing the
motion of a pendulum.

STUDENT A: Which of the lengths, AB or AC, should be con¬
sidered the deflection of the pendulum from the equilibrium
position (see Figure 50)? .

TEACHER: We are analysing the harmonic vibrations of a pendu¬
lum. For this it is necessary that the angle of maximum deviation

CAN YOU DEAL WITH HARMONIC VIBRATIONS?

125

of the string from the equilibrium position be very small

a « 1 (73)

(note that here angle a is expressed in radians; in degrees, angle
a should, in any case, be less than 10°). If condition (73) is com¬
plied with, the difference between the lengths AB and AC can be
neglected

AB = l sin a a* AC = l tan a

Thus your question becomes insignificant. For definiteness, we
can assume that x = AB = l sin a. Then equation (66) will take
the following form for a pendulum

mg s'ma = kl s'ma (66a)

from which

(74)

Substituting this equation into equation (72), we obtain the
formula for the period of harmonic vibrations of a pendulum

(75)

We shall also take up the question of the energy of the pendu¬
lum. Its total energy is evidently equal to mgh, where h is the
height to which the bob ascends at the extreme position (see
Figure 50). Thus

W = mgh

mgl (1 — cos a)

2 mgl

a

2

(76)

Relationship (76) is evidently suitable for all values of angle a. To
convert this result to relationship (68), it is necessary to satisfy
the condition of harmonicity of the pendulum’s vibrations, i.e.
inequality (73). Then, sin a can be approximated by the angle a
expressed in radians, and equation (76) will change to

W ss 2mgl

126 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Taking equation (74) into consideration, we finally obtain

W=k

as k

m 2

2

which is, in essence, the same as equation (68).

STUDENT B: If I remember correctly, in previously studying the
vibrations of a pendulum, there was generally no requirement
about the smallness of the angle of deviation.

TEACHER: This requirement is unnecessary if we only deal with
the energy of the bob or the tension of the string. In the given
case we are actually considering, not a pendulum, but the motion
of a ball in a circle in a vertical plane. However, if the prob¬
lem involves formula (75) for the period of vibrations, then
the vibration of the pendulum must necessarily be harmonic
and, consequently, the angle of deviation must be small. For
instance, in Problem No. 33, the condition of the smallness of
the angle of deviation of the pendulum is immaterial, while in
Problem No. 34 it is of vital importance.

Problems

32. A ball accomplishes harmonic vibrations as shown in Figure 48. Find
the ratio of the velocities of the ball at points whose distances from the
equilibrium position equal one half and one third of the amplitude.

33. A bob suspended on a string is deflected from the equilibrium position by
an angle of 60° and is then released. Find the ratio of the tensions of the
string for the equilibrium position and for the maximum deviation of the
bob.

34. A pendulum in the form of a ball (bob) on a thin string is deflected through
an angle of 5°. Find the velocity of the bob at the instant it passes the
equilibrium position if the circular frequency of vibration of the pendulum
equals 2/s.

§ 12 What Happens To A Pendulum In A State
Of Weightlessness?

TEACHER: Suppose we drive a nail in the wall of a lift and sus¬
pend a bob on a string of length / tied to the nail. Then we set
the bob into motion so that it accomplishes harmonic vibrations.
Assume that the lift ascends with an acceleration of a . What is
the period of vibration of the pendulum?

STUDENT A: When we go up in a lift travelling with acceleration,
we feel a certain increase in weight. Evidently, the pendulum
should “feel” the same increase. I think that its period of vibra¬
tion can be found by the formula

T=2n\A— (77)

V s + a

I cannot, however, substantiate this formula rigorously enough.

TEACHER: Your formula is correct. But to substantiate it we
will have to adopt a point of view that is new to us. So far we
have dealt with bodies located in inertial frames of reference
only, avoiding non-inertial frames. Moreover, I even warned you
against employing non-inertial frames of reference (§ 4). Be that
as it may, in the present section it is more convenient to use just
this frame of reference which, in the given case, is attached to the
accelerating lift.

Recall that in considering the motion of a body of mass m in
a non-inertial frame of reference having an acceleration a, we

128 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

must, on purely formal grounds, apply an additional force to the
body. This is called the force of inertia, equal to ma and acting
in the direction opposite to the acceleration. After the force of
inertia is applied to the body we can forget that the frame of
reference is travelling with acceleration, and treat the motion
as if it were in an inertial frame. In the case of the lift, we must
apply an additional force ma to the bob. This force is constant in
magnitude and its direction does not change and coincides with
that of the force of gravity mg. Thus it follows that in equation
(75) the acceleration g should be replaced by the arithmetical
sum of the accelerations (g + a). As a result, we obtain the
formula (77) proposed by you.

STUDENT A: Consequently, if the lift descends with a downward
acceleration a, the period of vibration will be determined by the
difference in the accelerations (g — a), since here the force of
inertia rna is opposite to the gravitational force. Is that correct?

TEACHER: Of course. In this case, the period of vibration of the
pendulum is

This formula makes sense on condition that a < g. The closer
the value of the acceleration a is to g, the greater the period of
vibration of the pendulum. At a = g, the state of weightlessness
sets in. What will happen to the pendulum in this case?

STUDENT A: According to formula (78), the period becomes
infinitely large. This must mean that the pendulum is stationary.

TEACHER: Let us clear up some details of your answer. We
started out with the pendulum vibrating in the lift. All of a
sudden, the lift breaks loose and begins falling freely downward
(we neglect air resistance). What happens to the pendulum?

STUDENT A: As I said before, the pendulum stops.

TEACHER: Your answer is not quite correct. The pendulum will
indeed be stationary (with respect to the lift, of course) if at the

WHAT HAPPENS TO A PENDULUM IN A STATE OF WEIGHTLESSNESS?

129

instant the lift broke loose the bob happened to be in one of
its extreme positions. If at that instant the bob was not at an
extreme position it will continue to rotate at the end of the string
in a vertical plane at a uniform velocity equal to its velocity at
the instant the accident happened.

STUDENT A: I understand now.

TEACHER: Then make a drawing illustrating the behaviour of a
pendulum (a bob attached to a string) inside a spaceship which is
in a state of weightlessness.

STUDENT A: In the spaceship, the bob at the end of the string
will either be at rest (with respect to the spaceship), or will ro¬
tate in a circle whose radius is determined by the length of the
string (if, of course, the walls or ceiling of the spaceship do not
interfere).

TEACHER: Your picture is not quite complete. Assume that we
are inside a spaceship in a state of weightlessness. We take the
bob and string and attach the free end of the string so that nei¬
ther walls nor ceiling interfere with the motion of the bob. After
this we carefully release the bob. The ball remains stationary.
Here we distinguish two cases: (1) the string is loose, and (2) the
string is taut.

Consider the first case (Position 1 in Figure 51 (a)). We impart a
certain velocity v Q to the bob. As a result, the bob will travel in
a straight line at uniform velocity until the string becomes taut
(Position 2 in Figure 51 (a)). At this instant, the reaction of the
string will act on the bob in the same manner as the reaction of
a wall acts on a ball bouncing off it. As a result, the direction
of travel of the bob will change abruptly and it will then again
travel at uniform velocity in a straight line (Position 3 in Fig¬
ure 51 (a)). In this peculiar form of “reflection” the rule of the
equality of the angles of incidence and reflection should be valid.

Now consider the second case: we first stretch the string taut and
then carefully release the bob. As in the first case, the bob will
remain stationary in the position it was released (Position 1 in

Figure 51: Anaysingthe
motion of a pendulum in a
spaceship.

130 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 51 (b)). Then we impart a certain velocity v 0 to the bob in
a direction perpendicular to the string. As a result the bob begins
to rotate in a circle at uniform velocity. The plane of rotation is
determined by the string and the vector of the velocity imparted
to the bob.

Let us consider the following problem. A string of length / with
a bob at one end is attached to a truck which slides without
friction down an inclined plane having an angle of inclination
a (Figure 52 (a)). We are to find the period of vibration of this
pendulum located in a frame of reference which travels with
a certain acceleration. However, in contrast to the preceding
problems with the lift, the acceleration of the system is at a
certain angle to the acceleration of the earth’s gravity. This poses
an additional question: what is the equilibrium direction of the
pendulum string?

STUDENT A: I once tried to analyse such a problem but became
confused and couldn’t solve it.

TEACHER: The period of vibration of a pendulum in this case is
found by formula (75) except that g is to be replaced by a certain
effective acceleration as in the case of the lift. This acceleration
(we shall denote it by g e jj) is equal to the vector sum of the
acceleration of gravity and that of the given system. Another
matter to be taken into account is that in the above mentioned
sum, the acceleration vector of the truck should appear with
the reversed sign, since the force of inertia is in the direction
opposite to the acceleration of the system. The acceleration
vectors are shown in Figure 52 (b), the acceleration of the truck
being equal to g sin a. Next we find g e jj

Se ff V S effx + S e

= \ (g sin a cos a) 2 + (g — gsin 2 a) 2 (^9)

^effy

= g cos a

from which

T = 2n

l

g cos a

(80)

Figure 52: Anaysing the
motion of a pendulum moving
with an accelerated frame at an
angle to earth’s gravity.

WHAT HAPPENS TO A PENDULUM IN A STATE OF WEIGHTLESSNESS? 131

STUDENT A: How can we determine the equilibrium direction of
the string?

TEACHER: It is the direction of the acceleration g e jr. On the
basis of equation (79) it is easy to see that this direction makes
an angle a with the vertical. In other words, in the equilibrium
position, the string of a pendulum on a truck sliding down an
inclined plane will be perpendicular to the plane.

STUDENT B: Isn’t it possible to obtain this last result in some
other way?

TEACHER: We can reach the same conclusion directly by consid¬
ering the equilibrium of the bob with respect to the truck. The
forces applied to the bob are: its weight mg, the tension T of
the string and the force of inertia ma (Figure 53). We denote the
angle the string makes with the vertical by (3.

Next we resolve all these forces in the vertical and horizontal
directions and then write the conditions of equilibrium for the
force components in each of these directions. Thus

T cos (3 + ma sin a = mg

T sin j3 = ma cos a

(81)

Taking into consideration that a = g sin a, we rewrite the system
of equations (81) in the form

T cos (3 = mg{ 1 — sin 2 a) )

T sin (3 = m g sin a cos a J

After dividing one equation by the other we obtain

cot /3 = cot a

Thus, angles /3 and a turn out to be equal. Consequently, the
equilibrium direction of the pendulum string is perpendicular to
the inclined plane.

STUDENT B: I have followed your explanations very closely and
come to the conclusion that I was not so wrong after all when, in

Figure 53: Anaysing the
motion of a pendulum.

132 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

answer to your question about the forces applied to a satellite, I
indicated the force of gravity and the centrifugal force (see § 8).

Simply, my answer should be referred to the frame of reference
attached to the satellite, and the centrifugal force is to be un¬
derstood as being the force of inertia. In a non-inertial frame of
reference attached to the satellite, we have a problem, not in dy¬
namics, but in statics. It is a problem of the equilibrium of forces
of which one is the centrifugal force of inertia.

TEACHER: Such an approach to the satellite problem is permissi¬
ble. However, in referring to the centrifugal force in § 8, you did
not consider it to be a force of inertia. You were simply trying to
think up something to keep the satellite from falling to the earth.
Moreover, in the case you mention, there was no necessity for
passing over to a frame of reference attached to the satellite: the
physical essence of the problem was more clearly demonstrated
without introducing a centrifugal force of inertia. My previous
advice is still valid: if there is no special need, do not employ a
non-inertial frame of reference.

The laws of statics are laws of equilibrium. Study these laws carefully.

Do not forget that they are of immense practical importance. A builder
without some knowledge of the basic laws of statics is inconceivable.
We shall consider examples illustrating the rules for the resolution of
forces. The subsequent discussion concerns the conditions of equilib¬
rium of bodies, which are used, in particular, for locating the centre of
gravity.

§ 13 Can You Use The Force Resolution Method
Efficiently?

TEACHER: In solving mechanical problems it is frequently nec¬
essary to resolve forces. Therefore, I think it would be useful
to discuss this question in somewhat more detail. First let us
recall the main rule: to resolve a force into any two directions it
is necessary to pass two straight lines through the head and two
more through the tail of the force vector, each pair of lines being
parallel to the respective directions of resolution. As a result we
obtain a parallelogram whose sides are the components of the
given force. This rule is illustrated in Figure 54 in which force F
is resolved in two directions: AA 1 and BB 1 .

Let us consider several problems in which force resolution is the
common approach. The first problem is illustrated in Figure 55:
we have two identical loads P suspended each from the middle
of a string. The strings sag due to the loads and make angles of
and a 2 with the horizontal. Which of the strings is subject to
greater tension?

STUDENT A: I can resolve the weight of each load on the same
drawing in directions parallel to the branches of the strings.
From this resolution it follows that the tension in the string is

2 sin a

Thus, the string which sags less is subject to greater tension.

Figure 54: Illustration for
resolution of forces resulting in
a parallelogram.

Figure 55: Anaysing the
motion of a pendulum.

136 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: Quite correct. Tell me, can we draw up the string so
tightly that it doesn’t sag at all when the load is applied?

STUDENT A: And why not?

TEACHER: Don’t hurry to answer. Make use of the result you
just obtained.

STUDENT A: Oh yes, I see. The string cannot be made so taut
that there is no sag. The tension in the string increases with a
decrease in angle a. However strong the string, it will be broken
by the tension when angle a becomes sufficiently small.

TEACHER: Note that the sagging of a string due to the action of
a suspended load results from the elastic properties of the string
causing its elongation. If the string could not deform (elongate)
no load could be hung from it. This shows that in construction
engineering, the strength analysis of various structures is closely
associated with their capability to undergo elastic deformations
(designers are wont to say that the structure must “breathe”).
Exceedingly rigid structures are unsuitable since the stresses
developed in them at small deformations may prove to be ex¬
cessively large and lead to failure. Such structures may even fail
under their own weight.

If we neglect the weight of the string in the preceding problem,
we can readily find the relationship between the angle a of sag
of the string and the weight P of the load. To do this we make
use of Hooke’s law for elastic stretching of a string or wire (see
Problem No. 35).

Consider another example. There’ is a Russian proverb, “a wedge
is driven out by a wedge” (the English equivalent being “like
cures like”). This can be demonstrated by applying the method
of force resolution (Figure 56 (a)). Wedge 1 is driven out of a
slot by driving wedge 2 into the same slot, applying the force F.
Angles a and j3 are given. Find the force that acts on wedge 1
and enables it to be driven out of the slot.

STUDENT A: I find it difficult to solve this problem.

CAN YOU USE THE FORCE RESOLUTION METHOD EFFICIENTLY? 137

TEACHER: Let us begin by resolving force F into components in
the horizontal direction and in a direction perpendicular to side
AB of wedge 2. The components obtained are denoted by F l and
F 2 (Figure 56 (b)). Component F 2 is counterbalanced by the reac¬
tion of the left wall of the slot; component F 1 equal to F/ tan a,
will act on wedge 1. Next we resolve this force into components
in the vertical direction and in a direction perpendicular to the
side CD of wedge 1. The respective components are S 3 and F 4
(Figure 56 (c)). Component F 4 is counterbalanced by the reaction
of the right wall of the slot, while component S 3 enables wedge
1 to be driven out of the slot. This is the force we are seeking. It
can readily be seen that it equals

Let us now consider a third example, illustrated in (Figure 57 (a)).

Two weights, P l and P 2 , are suspended from a string so that the
portion of the string between them is horizontal. Find angle (5
(angle a being known) and the tension in each portion of the string
(Tab> Trc and Tqy)).

This example resembles the preceding one with the wedges.

STUDENT A: First I shall resolve the weight P 4 into force com¬
ponents in the directions AB and PC (Figure 57 (b)). From this
resolution we find that

Tab ~

and T bc =

l

tana

Thus we have already found the tension in two portions of the
string. Next I shall resolve the weight P 2 into components in the
directions BC and CD (Figure 57 (c)). From this resolution we
can write the equations:

?bc ~

tan j3

2 and T cd =

P 2

sin j3

Equating the values for the tension in portion SC of the string
obtained in the two force resolutions, we can write

P 1

tana

Pi

tan f>

<«>

Figure 56: Anaysing the
motion of a pendulum.

(<*)

motion of a pendulum.

138 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

from which

/ P 2 tana

p = arctan

Substituting this value into the equation for T CD we can find the
tension in portion CD of the string.

TEACHER: Is it really so difficult to complete the problem, i.e. to
find the force T CD }

STUDENT A: The answer will contain the sine of the arctan (3,
i.e.

P.

Tcd =

sin ^arctan

P 2 tan a

TEACHER: Your answer is correct but it can be written in a
simpler form if sin j3 is expressed in terms of tan j3. As a matter
of fact

tan y?

Since

: obtain

sin j3 =

\J 1 + tan 2 13

tan j3 = tan a I —

T'cd ~

P^ t

tana

1 1 + ( j^\ tan 2 a

STUDENT B: I see that before taking an examination in physics,
you must review your mathematics very thoroughly.

TEACHER: Your remark is quite true.

Problems

35. An elastic string, stretched from wall to wall in a lift, sags due to the action
of a weight suspended from its middle point as shown in Figure 55. The
angle of sag a equals 30° when the lift is at rest and 45° when the lift travels
with acceleration. Find the magnitude and direction of acceleration of the
lift. The weight of the string is to be neglected.

CAN YOU USE THE FORCE RESOLUTION METHOD EFFICIENTLY? 139

36. A bob of mass m = 100 g is suspended from a string of length / = lm
tied to a bracket as shown in Figure 58 (a = 30°). A horizontal velocity of
2 m/s is imparted to the bob and it begins to vibrate as a pendulum. Find
the forces acting in members AB and B C when the bob is at the points of
maximum deviation from the equilibrium position.

Figure 58: Anaysing the mo¬
tion of a pendulum, Problem
36.

§ 14 What Do You Know About The Equilibrium
Of Bodies?

TEACHER: Two positions of equilibrium of a brick are shown
in Figure 59. Both equilibrium positions are stable, but their
degree of stability differs. Which of the two positions is the more
stable?

STUDENT A: Evidently, the position of the brick in Figure 59(a).

TEACHER: Why?

STUDENT A: Here the centre of gravity of the brick is nearer to
the earth’s surface,

TEACHER: This isn’t all.

(ay

(by

Figure 59: Which brick is more
stable?

STUDENT B: The area of the bearing surface is greater than in the
position shown in Figure 59 (b).

TEACHER: And this isn’t all either. To clear it up, let us consider
the equilibrium of two bodies: a rectangular parallelepiped with
a square base and a right circular cylinder Figure 60 (a). Assume
that the parallelepiped and cylinder are of the same height H
and have bases of the same area S. In this case, the centres of
gravity of the bodies are at the same height and, in addition, they
have bearing surfaces of the same area. Their degrees of stability,
however, are different.

142 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

The measure of the stability of a specific state of equilibrium is
the energy that must be expended to permanently disturb the
given state of the body.

Figure 60: Comparing equilib¬
rium of two bodies, which is
more stable?

STUDENT B: What do you mean by the word “permanently”?

TEACHER: It means that if the body is subsequently left to itself,
it cannot return to the initial state again. This amount of energy
is equal to the product of the weight of the body by the height
to which the centre of gravity must be raised so that the body
cannot return to its initial position. In the example with the
parallelepiped and cylinder, the radius of the cylinder is R =

a/ S/n and the side of the parallelepiped’s base is a = VS. To
disturb the equilibrium of the cylinder, its centre of gravity must

be raised through the height Figure 60 (b)

To disturb the equilibrium of the parallelepiped, its centre of

WHAT DO YOU KNOW ABOUT THE EQUILIBRIUM OF BODIES? 143

gravity must be raised Figure 60 (b)

it follows that h 2 < b x Thus, of the two bodies considered, the
cylinder is the more stable.

Now I propose that we return to the example with the two
positions of the brick.

STUDENT A: If we turn over the brick it will pass consecutively
from one equilibrium position to another. The dashed line in
Figure 61 shows the trajectory described by its centre of gravity
in this process. To change the position of a lying brick its centre
of gravity should be raised through the height h x expending
an energy equal to mgh x and to change its upright position,
the centre of gravity should be raised through h 2 , the energy
expended being mgh 2 . The greater degree of stability of the
lying brick is due to the fact that

mgh x > mgh 2 (82)

TEACHER: At last you’ve succeeded in substantiating the greater
stability of the lying position of a body.

STUDENT B: But it is evident that the heights h l and h 2 depend
upon the height of the centre of gravity above floor level and
on the area of the base. Doesn’t that mean that in discussing the
degree of stability of bodies it is correct to compare the heights
of the centres of gravity and the areas of the bases?

TEACHER: Why yes, it is, but only to the extent that these quan¬
tities influence the difference between the heights h 1 and h 2 .
Thus, in the example with the parallelepiped and cylinder, the
comparison of the heights of the centres of gravity and the areas
of the bases is insufficient evidence for deciding which of the

Figure 61: Trajectory described
by the centre of gravity of a
brick when turning over.

144 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

bodies is the more stable. Besides, I wish to draw your attention
to the following. Up till now we have tacitly assumed that the
bodies were made of the same material. In this case, the inequal¬
ity (82) could be satisfied by observing the geometric condition

h x > h 2 .

In the general case, however, bodies may be made of different
materials, and the inequality (82) may be met even when h j < h 2
owing to the different densities of the bodies. For example, a
cork brick will be less stable in the lying position than a lead
brick in the upright position. Let us now see what conditions for
the equilibrium of bodies you know.

STUDENT A: The sum of all the forces applied to a body should
equal zero. In addition, the weight vector of the body should fall
within the limits of its base.

TEACHER: Good. It is better, however, to specify the conditions
of equilibrium in a different form, more general and more con¬
venient for practical application. Distinction should be made
between two conditions of equilibrium:

First condition: The projections of all forces applied to the body
onto any direction, should mutually compensate one an¬
other. In other words, the algebraic sum of the projections
of all the forces onto any direction should equal zero.

This condition enables as many equations to be writ¬
ten as there are independent directions in the problem:
one equation for a one-dimensional problem, two for a
two-dimensional problem and three for the general case
(mutually perpendicular directions are chosen).

Second condition (moment condition): The algebraic sum of the
moments of the forces about any point should equal zero.
Here, all the force moments tending to turn the body
about the chosen point in one direction (say, clockwise)
are taken with a plus sign and all those tending to turn
the body in the opposite direction (counterclockwise)
are taken with a minus sign. To specify the moment
condition, do the following:

WHAT DO YOU KNOW ABOUT THE EQUILIBRIUM OF BODIES?

145

(a) establish all forces applied to the body;

(b) choose a point with respect to which the force
moments are to be considered;

(c) find the moments of all the forces with respect to
the chosen point;

(d) write the equation for the algebraic sum of the
moments, equating it to zero.

In applying the moment condition, the following should be kept
in mind:

(1) the above stated condition refers to the case when all the
forces in the problem and their arms are in a single plane
(the problem is not three-dimensional), and

(2) the algebraic sum of the moments should be equated to
zero with respect to any point, either within or outside
the body.

It should be emphasized that though the values of the separate
force moments do depend upon the choice of the point - with
respect to which the force moments are considered), the algebraic
sum of the moments equals zero in any case. To better under¬
stand the conditions of equilibrium, we shall consider a specific
problem.

A beam of weight P x is fixed at points B and C (Figure 62 (a)). At
point D, a load with a weight of P 2 is suspended from the beam.
The distances AS = a,BC = 2a and CD = a. Find the reactions
N b and N c at the two supports. Assume that the reactions of the
supports are directed vertically.

As usual, first indicate the forces applied to the body.

STUDENT A: The body in the given problem is the beam. Four
forces are applied to it: weights P 1 and P 2 and reactions Ng and

A’c-

te ac 11 ER: Indicate these forces on the drawing.

146

QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 62: A beam with
suspended loads, the problem
is to find the reactions of the
supports.

STUDENT A: But I don’t know whether the reactions are directed
upward or downward.

TEACHER: Assume that both reactions are directed upward.

STUDENT A: Well, here is my drawing (Figure 62 (b)). Next I can
specify the first condition of equilibrium by writing the equation

N b + N c = P 1 + Pi

TEACHER: I have no objection to this equation as such. However,
in our problem it is simpler to use the second condition of equi¬
librium (the moment condition), employing it first with respect
to point B and then to point C.

STUDENT A: All right, I’ll do just that. As a result I can write the

WHAT DO YOU KNOW ABOUT THE EQUILIBRIUM OF BODIES?

147

equations

with respect to point B : aP l — 2 aN c + 3aP 2 = 0
with respect to point C: 2 aN B — aP [ + aP 2 = 0

(83)

TEACHER: Now you see: each of your equations contains only
one of the unknowns. It can readily be found.

STUDENT A: From equations (83) we find

N b =

N b =

(Pi-Pi)
2

(Pi + 3P 2 )

(84)

(85)

TEACHER: Equation (85) always has a positive result. This means
that reaction N c is always directed upward (as we assumed).
Equation (84) gives a positive result when > P 2 , negative when
P j < P 2 and becomes zero when P j = P-,. This means that when
P j < Pt, reaction N B is in the direction we assumed, i. e. upward
(see (Figure 62 (b)); that when P l < P 2 , reaction N B is downward
(see (Figure 62 (c)); and at P 1 = P 2 there is no reaction N B .

§75 How Do You Locate The Centre Of Gravity ?

TEACHER: In many cases, examinees find it difficult to locate the
centre of gravity of a body of system of bodies. Is everything
quite clear to you on this matter?

STUDENT A: No, I can’t say it is. I don’t quite understand how
you find the centre of gravity in the two cases shown in Fig¬
ure 63 (a) and Figure 64 (a).

Figure 63: Problem is to find
the centre of gravity of the
given body.

TEACHER: All right. In the first case it is convenient to divide

150 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

the plate into two rectangles as shown by the dashed line in
Figure 63 (b). The centre of gravity of rectangle 1 is at point
A; the weight of this rectangle is proportional to its area and is
equal, as is evident from the figure, to 6 units (here the weight
is conditionally measured in square centimetres). The centre
of gravity of rectangle 2 is at point the weight of this rectangle
is equal to 10 units. Next we project the points A and B on the
coordinate axes O x and O y ; these projections are denoted byA l
and B l on the X-axis and by A 2 and B 2 on the F-axis. Then we
consider the “bars” A { B l and A 2 B 2 , assuming that the masses are
concentrated at the ends of the “bars”, the mass of each end being
equal to that of the corresponding rectangle (see Figure 63 (b).

As a result, the problem of locating the centre of gravity of our
plate is reduced to finding the centres of gravity of “bars” A l B l
and A 2 B 2 The positions of these centres of gravity will be the
coordinates of the centre of gravity of the plate.

But let us complete the problem. First we determine the location
of the centre of gravity of “bar” AjSj using the well-known rule
of force moments (see Figure 63 (b)):

6x = 10(2 — x) then, x = - cm

Thus, the X-coordinate of the centre of gravity of the plate in the
chosen system of coordinates is

9

X = (1 + x) cm = - cm
V ' 4

In a similar way we find the centre of gravity of “bar” A 2 B 2 :

6 y = 10(1 —y)

from which it follows that y = - cm. Thus the F-coordinate of

8 .

the centre of gravity of the plate is

17

Y = (1.5 + y)cm = —cm

STUDENT A: Now I understand. That is precisely how I would
go about finding coordinate X of the centre of gravity of the

HOW DO YOU LOCATE THE CENTRE OF GRAVITY? 151

plate. I was not sure that coordinate Y could be found in the
same way.

TEACHER: Let us consider the second case, shown in Figure 64 (a).

Figure 64: Problem is to find
the centre of gravity of the
given body.

Two approaches are available. For instance, instead of the given
circle with one circular hole, we can deal with a system of two
bodies: a circle with two symmetrical circular holes and a cir¬
cle inserted into one of the holes (Figure 64 (b)). The centres
of gravity of these bodies are located at their geometric centres.
Knowing that the weight of the circle with two holes is propor¬
tional to its area, i.e.

2nR 2 \ nR 2
~)

and that of the small circle is proportional to its area nR 2 / 4, we
reduce the problem to finding the point of application of the
resultant of the two parallel forces shown below in Figure 64 (b).
We denote by x the distance from the sought-for centre of grav¬
ity to the geometric centre of the large circle. Then, according to

152 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 64 (b), we can write

n R 2

~Y~

x from which x

R

6

There is another possible approach. The given circle with the
hole can be replaced by a solid circle (with no hole) plus a circle
located at the same place where the hole was and having a neg¬
ative weight (i.e. one acting upward) (Figure 64 (c)) which will
compensate for the positive weight of the corresponding portion
of the solid circle. As a whole, this arrangement corresponds to
the initial circle with the circular hole.

In this case, the problem is again reduced to finding the point
of application of the resultant of the two forces shown at the
bottom of Figure 64 (c). According to the diagram we can write:

7iR 2 x =

from which, as in the preceding case, x

R

6 '

STUDENT A: I like the first approach better because it does not
require the introduction of a negative weight.

TEACHER: In addition, I want to propose a problem involving
locating the centre of gravity of the system of loads shown
in Figure 65 (a). We are given six loads of different weights
(P l ,P 2 , • • ■ > P(,) arranged along a bar at equal distances a from
one another. The weight of the bar is neglected. How would you
go about solving this problem?

STUDENT A: First I would consider two loads, for instance,

P l and P 2 , and find the point of application of their resultant.
Then I would indicate this resultant (equal to the sum P l 4- P 2 )
on the drawing and would cross out forces P 1 and P 2 , from
further consideration. Now, instead of the six forces, only five
would remain. Next, I would find the point of application of
the resultant of another pair of forces, etc. Thus, by consecutive
operations I would ultimately find the required resultant whose
point of application is the centre of gravity of the whole system.

HOW DO YOU LOCATE THE CENTRE OF GRAVITY? 153

TEACHER: Though your method of solution is absolutely cor¬
rect, it is by far too cumbersome. I can show you a much more
elegant solution. We begin by assuming that we are sup porting
the system at its centre of gravity (at point B in Figure 65 (b).

STUDENT B: (interrupting): But you don’t yet know the location
of the centre of gravity. How do you know that it is between the
points of application of forces P 3 and P 4 }

TEACHER: It makes no difference to me where exactly the cen¬
tre of gravity is. I shall not take advantage of the fact that in
Figure 65 (b) the centre of gravity turned out to be between the
points of application of forces P 3 and P 4 . So we assume we are
supporting the system at its centre of gravity. As a result, the
bar is in a state of equilibrium. In addition to the six forces, one
more force - the bearing reaction N - will act on the bar. Since
the bar is in a state of equilibrium, we can apply the conditions
of equilibrium (see § 14). We begin with the first condition of

Figure 65: Problem is to
find the centre of gravity
of the bodies in the given
configuration.

154 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

equilibrium for the projection of all the forces in the vertical
direction

N = P l +P 2 + P 3 +P 4 + P 5 +P (> (86)

Then we apply the second condition (moment condition),
considering the force moments with respect to point A in Fig¬
ure 65 (b) (i.e. the left end of the bar). Here, all the forces tend to
turn the bar clockwise, and the bearing reaction tends to turn it
counterclockwise. We can write

N (AB ) = aP 2 + 2 aP 3 + 3 aP 4 + 4aP 5 + 5aP b (87)

Combining conditions (86) and (87), we can find the length AB,
i.e. the required position of the centre of gravity measured from
the left end of the bar

^ ^ aP 2 4- 2aP 2 + 3aP 4 4- 4 aP^ 4- 5 aP^
P 1 +P 2 +P 3 + P 4 +P 5 + P 6

( 88 )

STUDENT A: Yes, I must admit that your method is much sim¬
pler.

TEACHER: Also note that your method of solving the problem
is very sensitive to the number of loads on the bar (the addition
of each load makes the solution more and more tedious). My
solution, on the contrary, does not become more complicated
when loads are added. With each new load, only one term is
added to the numerator and one to the denominator in equation
( 88 ).

STUDENT B: Can we find the location of the centre of gravity of
the bar if only the moment condition is used?

TEACHER: Yes, we can. This is done by writing the condition of
the equilibrium of force moments with respect to two different
points. Let us do precisely that. We will consider the condition
for the force moments with respect to points A and C (see Fig¬
ure 65 (b)). For point A the moment condition is expressed by
equation (87); for point C, the equation will be

N(5a —AB) = aP 5 4- 2aP 4 4- 3aP 2 + 4aP 2 4- 5aP x (89)

HOW DO YOU LOCATE THE CENTRE OF GRAVITY? 155

Dividing equation (87) by (89) we obtain

AB aP 2 + 2a P-^ ~T 3 uP^ + 4 aP^ -T 5aP^

5 a —AB aP^2aP 4 3aP^4aP 2 5aP^

From which

AB {aP ^ F 2aP 4 + 3 aP^ H- 4aP 2 4- 5 aP^ H-
<*P 2 + 2a P 3 H- 3 aP 4 + AaP 3 + 5 aP^j
= 5a (aP 2 + 2aP 3 + 3^P 4 + 4aP 5 + 5 aP G )

or

AB X 5<2 (P | 4~ P 2 + P3 + P4 + P5 -h P^)

= 5^ (aP 2 + 2^P 3 + 3^P 4 + 4*2 P 5 + 5^P 6 )

Thus we obtain the same result as in equation (88).

Problems

37. Locate the centre of gravity of a circular disk having two circular
holes as shown in Figure 66. The radii of the holes are equal to one
half and one fourth of the radius of the disk.

Figure 66: Problem is to find
the centre of gravity of the
given body.

Archimedes' principle does not usually draw special attention. This is a
common mistake of students preparing for physics exams. Highly inter¬
esting questions and problems can be devised on the basis of this prin¬
ciple. We shall discuss the problem of the applicability of Archimedes'
principle to bodies in a state of weightlessness.

% 16 Do you know Archimedes'principle

TEACHER: Do you know Archimedes’ principle?

STUDENT A: Yes, of course. The buoyant force exerted by a
liquid on a body immersed in it is exactly equal to the weight of
the liquid displaced by the body.

TEACHER: Correct. Only it should be extended to include gases:
a buoyant force is also exerted by a gas on a body “immersed” in
it. And now can you give a theoretical proof of your statement?

STUDENT A: A proof of Archimedes’ principle?

TEACHER: Yes.

STUDENT A: But Archimedes’ principle was discovered directly
as the result of experiment.

TEACHER: Quite true. It can, however, be derived from simple
energy considerations. Imagine that you raise a body of volume
V and density p to a height H, first in a vacuum and then in a
liquid with a density p 0 . The energy required in the first case
equals pgVH. The energy required in the second case is less
because the raising of a body of volume V by a height H is
accompanied by the lowering of a volume V of p the liquid
by the same height H. Therefore, the energy expended in the
second case equals

PgVH - p 0 gVH

Regarding the subtrahend p 0 g VH as the work done by a certain
force, we can conclude that, compared with a vacuum, in a liquid

160 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

an additional force F = p 0 gVH acts on the body making it
easier to raise. This force is called the buoyant force. Quite obvi¬
ously, it is exactly equal to the weight of the liquid in the volume
V of the body immersed in the liquid. (Note that we neglect the
energy losses associated with friction upon real displacements of
the body in the liquid.)

P Figure 67: Buoyant force for

Archimedes’ principle.

P + Pogh

Archimedes’ principle can be deduced in a somewhat different
way. Assume that the body immersed in the liquid has the form
of a cylinder of height h and that the area of its base is S (Fig¬
ure 67). Assume that the pressure on the upper base is p.

Then the pressure on the lower base will equal p + pgh. Thus,
the difference in pressure on the upper and lower bases equals
Pogb. If we multiply this difference by the area S of the base, we
obtain the force F = pgbS which tends to push the body up¬
ward. Since hS = V, the volume of the cylinder, it can readily be
seen that this is the buoyant force which appears in Archimedes’
principle.

STUDENT A: Yes, now I see that Archimedes’ principle can be
arrived at by purely logical reasoning.

TEACHER: Before proceeding any further, let us recall the condi¬
tion for the floating of a body.

STUDENT A: I remember that condition. The weight of the body
should be counterbalanced by the buoyant force acting on the
body in accordance with Archimedes’ principle.

DO YOU KNOW ARCHIMEDES’ PRINCIPLE?

161

TEACHER: Quite correct. Here is an example for you. A piece of
ice floats in a vessel with water. Will the water level change when
the ice melts?

STUDENT A: The level will remain unchanged because the weight
of the ice is counterbalanced by the buoyant force and is there¬
fore equal to the weight of the water displaced by the ice. When
the ice melts it converts into water whose volume is equal to that
of the water that was displaced previously.

TEACHER: Exactly. And now let us assume that there is, for
instance, a piece of lead inside the ice. What will happen to the
water level after the ice melts in this case?

STUDENT A: I’m not quite sure, but I think the water level
should reduce slightly. I cannot, however, prove this.

TEACHER: Let us denote the volume of the piece of ice together
with the lead by V, the volume of the piece of lead by v, the
volume of the water displaced by the submerged part of the ice
by V l the density of the water by p 0 , the density of the ice by p x
and the density of the lead by p 2 . The piece of ice together with
the lead has a weight equal to

p l g(V-v)+p 2 gv

This weight is counterbalanced by the buoyant force pogV t.
Thus

P\g(V -v)+ p 2 gv = p 0 gVt (90)

After melting, the ice turns into water whose volume V 9 is found
from the equation

Pig(V~v) =p 0 gV 2

Substituting this equation into (90) we obtain

Po 8 V 2 +P 2 gv = PoS V i

From which we find that the volume of water obtained as a result
of the melting of the ice is

V, = Vi-v^

Pi

(91)

162 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Thus, before the ice melted, the volume of water displaced was
Vj. Then the lead and the water from the melted ice began to
occupy the volume ( V 2 + v). To answer the question concerning
the water level in the vessel, these volumes should be compared.
From equation (91) we get

V 2 + v=V l -v P2 ~ Po (92)

P o

Since p 2 > Po (lead is heavier than water), it can be seen from
equation (92) that ( V 2 + v) < V l . Consequently, the water level
will reduce as a result of the melting of the ice. Dividing the
difference in the volumes V l — ( V 2 + v) by the cross-sectional
area S of the vessel (assuming, for the sake of simplicity, that it is
of cylindrical shape) we can find the height h by which the level
drops after the ice melts. Thus

(95)

Do you understand the solution of this problem?

STUDENT A: Yes, I’m quite sure I do.

TEACHER: Then, instead of the piece of lead, let us put a piece
of cork of volume v and density p 2 inside the ice. What will
happen to the water level when the ice melts?

STUDENT A: I think it will rise slightly.

TEACHER: Why?

STUDENT A: In the example with lead the level fell. Lead is
heavier than water, and cork is lighter than water. Consequently,
in the case of cork we should expect the opposite effect: the
water level should rise.

TEACHER: You are mistaken. Your answer would be correct if
the cork remained submerged after the ice melted. Since the cork
is lighter than water it will surely rise to the surface and float.
Therefore, the example with cork (or any other body lighter
than water) requires special consideration. Using the result of

DO YOU KNOW ARCHIMEDES’ PRINCIPLE? 163

equation (91), we can find the difference between the volume of
the water displaced by the piece of ice together with the cork,
and that of the water obtained by the melting of the ice. Thus

(94)

Next we apply the condition for the floating of the piece of cork:

(95)

p 3 v=p Q v l

where v 1 is the volume of the part of the cork submerged in
water. Substituting this equation into (94), we obtain

Vi — V? + zq

Thus the volume of water displaced by the piece of ice is ex¬

actly equal to the sum of the volume of water obtained from

the melted ice. and the volume displaced by the submerged por¬
tion of the floating piece of cork. So in this case the water level
remains unchanged.

STUDENT A: And if the piece of ice contained simply a bubble of
air instead of the piece of cork?

TEACHER: After the ice melts, this bubble will be released. It can
readily be seen that the water level in the vessel will be exactly
the same as it was before the ice melted. In short, the example
with the bubble of air in the ice is similar to that with the piece
of cork.

STUDENT A: I see that quite interesting question and problems
can be devised on the basis of Archimedes’ principle.

TEACHER: Unfortunately, some examinees don’t give enough
attention to this principle when preparing for their physics ex¬
aminations. Let us consider the following example. One pan
of a balance carries a vessel with water and the other, a stand
with a weight suspended from it. The pans are balanced (Fig¬
ure 68 (a)). Then the stand is turned so that the suspended weight
is completely submerged in the water. Obviously, the state of
equilibrium is disturbed since the pan with the stand becomes

164 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

lighter (Figure 68 (b)). What additional weight must be put on
the pan with the stand to restore equilibrium?

Figure 68: What additional
weight must be put on the
pan with the stand to restore
equilibrium?

STUDENT A: The submerged weight is subject to a buoyant
force equal to the weight of the water of the volume displaced
by the submerged weight (we denote this weight of water by
P). Consequently, to restore equilibrium, a weight P should be
placed on the pan with the stand.

TEACHER: You are mistaken. You would do well to recall New¬
ton’s third law of motion. According to this law, the force with
which the water in the vessel acts on the submerged weight is ex¬
actly equal to the force with which the submerged weight acts on
the water in the opposite direction. Consequently, as the weight
of the pan with the stand reduces, the weight of the pan with the
vessel increases. Therefore, to restore equilibrium, a weight equal
to 2 P should be added to the pan with the stand.

STUDENT A: I can’t quite understand your reasoning. After all,
the interaction of the submerged weight and the water in no way
resembles the interaction of two bodies in mechanics.

TEACHER: The field of application of Newton’s third law is not
limited to mechanics. The expression “to every action there is an
equal and opposite reaction” refers to a great many kinds of in¬
teraction. We can, however, apply a different line of reasoning in
our case, one to which you will surely have no objections. Let us

DO YOU KNOW ARCHIMEDES’ PRINCIPLE?

165

deal with the stand with the weight and the vessel with the water
as part of a single system whose total weight is obviously the sum
of the weight of the left pan and that of the right pan. The total
weight of the system should not change due to interaction of its
parts with one another. Hence, if as the result of interaction the
weight of the right pan is decreased by P, the weight of the left
pan must be increased by the same amount (P). Therefore, after
the weight is submerged in the vessel with water, the difference
between the weights of the left and right pans should be 2 P.

Problems

38. A vessel of cylindrical shape with a cross-sectional area S is filled
with water in which a piece of ice, containing a lead ball, floats. The
volume of the ice together with the lead ball is V and 1/20 of this
volume is above the water level. To what mark will the water level
in the vessel reduce after the ice melts? The densities of water, ice
and lead are assumed to be known.

§ 17 Is Archimedes' Principle Valid In A Space¬
ship?

TEACHER: Is Archimedes’ principle valid in a spaceship when it
is in a state of weightlessness?

STUDENT A: I think it is not. The essence of Archimedes’ prin¬
ciple is that due to the different densities of the body and the
liquid (of equal volumes, of course), different amounts of work
are required to raise them to the same height. In a state of weight¬
lessness, there is no difference in these amounts of work since
the work required to lift a body and that required to lift an equal
volume of the liquid is equal to zero.

We can reach the same conclusion if we consider the pressure of
the liquid on a body submerged in it because the buoyant force is
due to the difference in the pressures exerted on the bottom and
top bases on the body. In a state of weightlessness, this difference
in pressure vanishes and, with it, the buoyant force. I may add
that in a state of weightlessness there is no difference between
“up” and “down” and so it is impossible to indicate which base of
the body is the upper and which the lower one.

Thus, in a state of weightlessness, no buoyant force acts on a
body submerged in a liquid. This means that Archimedes’ princi¬
ple is not valid for such a state.

STUDENT B: I don’t agree with the final conclusion of STU¬
DENT A. I am sure that Archimedes’ principle is valid for a state

168 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

of weightlessness. Let us reason more carefully. We shall not
pass over directly to a state of weightlessness, but begin with a
lift travelling with a certain acceleration a which is in the same
direction as the acceleration g of gravity. Assume that a < g. It
is easy to see that in the given case a body submerged in a liquid
will be subject to the buoyant force

F = Po(g~ a ) v ( % )

and the weight of the liquid of a volume displaced by the body is
also equal to p 0 (g — a) V. Thus, the buoyant force is still equal to
the weight of the liquid displaced by the body, i.e. Archimedes’
principle is valid. Next we will gradually increase the accelera¬
tion a, approaching the value of g. According to equation (96),
the buoyant force will be gradually reduced, but simultaneously
and in exactly the same way, the weight of a volume of liquid
equal to the volume of the body will also be reduced. In other
words, as acceleration a approaches acceleration g, Archimedes’
principle will continue to be valid. In the limit a = g a state of
weightlessness sets in. At this the buoyant force becomes zero,
but so does the weight of the liquid displaced by the body. Con¬
sequently, nothing prevents us from stating that Archimedes’
principle is valid for a state of weightlessness as well. I wish to
illustrate my argument by the following example. Let us suppose
that a piece of cork floats in a vessel with water. According to
equation (95) the ratio of the volume of the piece of cork sub¬
merged in the water to the total volume of the piece is equal to
the ratio of the density of cork to the density of water. Thus

= Pi
v Po

(97)

Next, we suppose that this vessel is in a lift and the lift begins to
descend with a certain acceleration a. Since this does not change
the densities of cork and water, equation (97) holds. In other
words, in the motion of the lift with acceleration, the position
of the piece of cork with reference to the water level remains the
same as in the absence of acceleration. Obviously, this condition
will not change in the limiting case when a = g and we reach a
state of weightlessness. In this way, the position of the piece of

IS ARCHIMEDES’ PRINCIPLE VALID IN A SPACESHIP?

169

cork with respect to the water level, determined by Archimedes’
principle, turns out to be independent of the acceleration of the
lift. In this case no distinction can be made between the presence
and absence of weightlessness.

TEACHER: I should say that both of your arguments are well sub¬
stantiated. However, I must agree with Student A: Archimedes’
principle is not valid for a state of weightlessness.

STUDENT B: But then you must refute my proofs.

TEACHER: That’s just what I’ll try to do. Your arguments are
based on two main points. The first is that at an acceleration a <
g a body is buoyed up in the liquid in a manner fully complying
with Archimedes’ principle. The second is that this statement
must hold for the limiting case as well, when a = g, i.e. a state of
weightlessness is reached. I have no objection to the first point,
but I don’t agree with the second.

STUDENT B: But you can’t deny that the piece of cork remains
in the same position in a state of weightlessness as well! And its
position directly follows from Archimedes’ principle.

TEACHER: Yes, that’s true. The piece of cork actually does re¬
main in the same position in a state of weightlessness as well.
However, in this state its position with respect to the surface of
the liquid is no longer a result of Archimedes’ principle. Push
it deep into the water with your finger and it will remain sus¬
pended at the depth you left it. On the other hand, if there
is even the smallest difference (g — a) , the piece of cork will
come up to the surface and float in the position determined by
Archimedes’ principle. Thus, there is a basic difference between
weightlessness and the presence of even an insignificant weight-
ness. In other words, in passing over to a state of weightlessness,
at the “very last instant” there occurs an abrupt change, or jump,
that alters the whole situation qualitatively.

STUDENT B: But what is this jump due to? Where did it come
from? In my reasoning, acceleration a smoothly approached
acceleration g.

170 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: This jump is related to the fact that at a = g, a certain
symmetry appears: the difference between “up” and “down”
disappears, which, incidentally, was very aptly pointed out by
Student A. If the difference (g — a) is infinitely small, but still
not equal to zero, the problem contains a physically defined
direction “upward”. It is precisely in this direction that the buoy¬
ant force acts. However, at a = g, this direction disappears, and
all directions become physically equivalent. That’s what I mean
by a jump. The destruction or the appearance of symmetry
always occurs with jump.

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Basically, modern physics is molecular physics. Hence it is especially im¬
portant to obtain some knowledge of the fundamentals of the molecular-
kinetic theory of matter, if only by using the simplest example of the
ideal gas. The question of the peculiarity in the thermal expansion of
water is discussed separately. The gas laws will be analysed in detail
and will be applied in the solution of specific engineering problems.

§ 18 What Do You Know About The Molecular-
Kinetic Theory Of Matter?

TEACHER: One of the common examination questions is: what
are the basic principles of the molecular-kinetic theory of matter?
How would you answer this question?

STUDENT A: I would mention the two basic principles. The first
is that all bodies consist of molecules, and the second, that the
molecules are in a state of chaotic thermal motion.

TEACHER: Your answer is very typical: laconic and quite incom¬
plete. I have noticed that students usually take a formal attitude
with respect to this question. As a rule, they do not know what
should be said about the basic principles of the molecular-kinetic
theory, and explain it away with just a few general remarks. In
this connection, I feel that the molecular-kinetic theory of matter
should be discussed in more detail. I shall begin by mentioning
the principles of this theory that can be regarded as the basic
ones.

1. Matter has a “granular” structure: it consists of molecules
(or atoms). One gram-molecule of a substance contains
N a = 6 x 10 23 molecules regardless of the physical state
of the substance (the number N A is called Avogadro’s
number).

2. The molecules of a substance are in a state of incessant
thermal motion.

174 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

3. The nature of the thermal motion of the molecules de¬
pends upon the nature of their interaction and changes
when the substance goes over from one physical state to
another.

4. The intensity of the thermal motion of the molecules
depends upon the degree to which the body is heated,
this being characterized by the absolute temperature T.
The theory proves that the mean energy e of a separate
molecule is proportional to the temperature T. Thus, for
instance, for monoatomic (single-atom) moleculese

e = \kT (98)

where k = 1.38 x 10~ 16 erg/deg is a physical constant
called Boltzmann’s constant.

5. From the standpoint of the molecular-kinetic theory, the
total energy £ of a body is the sum of the following terms:

E = E k +E p + U (99)

where E^ is the kinetic energy of the body as a whole, E p
is the potential energy of the body as a whole in a certain
external field, and U is the energy associated with the
thermal motion of the molecules of the body. Energy U
is called the internal energy of the body. Inclusion of the
internal energy in dealing with various energy balances is
a characteristic feature of the molecular-kinetic theory.

STUDENT B: We are used to thinking that the gram-molecule and
Avogadro’s number refer to chemistry.

TEACHER: Evidently, that is why students taking a physics exam¬
ination do not frequently know what a gram-molecule is, and,
as a rule, are always sure that Avogadro’s number refers only to
gases. Remember: a gram-molecule is the number of grams of
a substance which is numerically equal to its molecular weight
(and by no means the weight of the molecule expressed in grams,
as some students say); the gram-atom is the number of grams of a
substance numerically equal to its atomic weight; and Avogadro’s

WHAT DO YOU KNOW ABOUT THE MOLECULAR-KINETIC THEORY OF MATTER? 175

number is the number of molecules in a gram-molecule (or atoms
in a gram-atom) of any substance, regardless of its physical state.

I want to point out that Avogadro’s number is a kind of a bridge
between the macro and micro characteristics of a substance.

Thus, for example, using Avogadro’s number, you can express
such a micro characteristic of a substance as the mean distance
between its molecules (or atoms) in terms of the density and
molecular (or atomic) weight. For instance, let us consider solid
iron. Its density is p = 7.8 g/cm 3 and atomic weight A = 56. We
are to find the mean distance between the atoms in iron. We shall
proceed as follows: in A g of iron there are N A atoms, then in 1 g
of iron there must be N a /A atoms. It follows that in 1 cm 3 there
are pN A /A atoms. Thus each atom of iron is associated with a
volume of A/pN A crc?. The required mean distance between the
atoms is approximately equal to the cube root of this volume

STUDENT B: Just before this you said that the nature of the ther¬
mal motion of the molecules depends upon the intermolecular
interaction and is changed in passing over from one physical state
to another. Explain this in more detail, please.

TEACHER: Qualitatively, the interaction of two molecules can
be described by means of the curve illustrated in Figure 69. This
curve shows the dependence of the potential energy E^ of inter¬
action of the molecules on the distance r between their centres.
At a sufficiently large distance between the molecules the curve
Ep(r) asymptotically approaches zero, i.e. the molecules practi¬
cally cease to interact. As the molecules come closer together, the
curve Ep(r) turns downward. Then, when they are sufficiently
close to one another, the molecules begin to repulse one another
and curve Ep(r) turns upward and E continues to rise (this re¬
pulsion means that the molecules cannot freely penetrate into
each other). As can be seen, the Ep{r) curve has a characteristic
minimum.

Figure 69: Dependence of
the potential energy of
interaction of the molecules on
the distance r between their
centres.

STUDENT B: What is negative energy?

176 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: As we know, energy can be measured from any value.
For instance, we can measure the potential energy of a stone
from ground level of the given locality, or we can measure it
from sea level, it makes no difference. In the given case, the
zero point corresponds to the energy of interaction between
molecules separated from each other at an infinitely large dis¬
tance. Therefore, the negative energy of the molecule means that
it is in a bound state (bound with another molecule). To “free”
this molecule, it is necessary to add some energy to it to increase
the energy of the molecule to the zero level. Assume that the
molecule has a negative energy e 1 (see Figure 69). It is evident
from the curve that in this case the molecule cannot get farther
away from its neighbour than point B or get closer than point A.
In other words, the molecule will vibrate between points A and B
in the field of the neighbouring molecule (more precisely, there
will be relative vibration of two molecules forming a bound
system).

In a gas molecules are at such great distances from one another
on an average that they can be regarded as practically nonin¬
teracting. Each molecule travels freely, with relatively rare col¬
lisions. Each molecule participates in three types of motion:
translatory, rotary (the molecule rotates about its own axis) and
vibratory (the atoms in the molecule vibrate with respect to one
another). If a molecule is monoatomic it will have only transla¬
tory motion.

In a crystal the molecules are so close together that they form
a single bound system. In this case, each molecule vibrates in
some kind of general force field set up by the interaction of the
whole collective of molecules. Typical of a crystal as a common
bound system of molecules is the existence of an ordered three-
dimensional structure - the crystal lattice. The lattice points
are the equilibrium positions of the separate molecules. The
molecules accomplish their complex vibratory motions about
these positions. It should be noted that in some cases when
molecules form a crystal, they continue to retain their individ¬
uality to some extent. In these cases, distinction is to be made
between the vibration of the molecule in the field of the crystal

WHAT DO YOU KNOW ABOUT THE MOLECULAR-KINETIC THEORY OF MATTER?

177

and the vibration of the atoms in the separate molecules. This
phenomenon occurs when the binding energy of the atoms in
the molecules is substantially higher than the binding energy
of the molecules themselves in the crystal lattice. In most cases,
however, the molecules do not retain their individuality upon
forming a crystal so that the crystal turns out to be made up, not
of separate molecules, but of separate atoms. Here, evidently,
there is no intramolecular vibration, but only the vibration
of the atoms in the field of the crystal. This, then, is the min¬
imum amount of information that examinees should possess
about atomic and molecular thermal motions in matter. Usually,
when speaking about the nature of thermal motions in matter,
examinees get no farther than saying it is a “chaotic motion”,
thus trying to cover up the lack of more detailed knowledge of
thermal motion.

STUDENT B: But you haven’t said anything about the nature of
the thermal motions of molecules in a liquid.

TEACHER: Thermal motions in a liquid are more involved than
in other substances. A liquid occupying an intermediate position
between gases and crystals exhibits, along with strong particle
interaction, a considerable degree of disorder in its structure.

The difficulty of dealing with crystals, owing to the strong inter¬
action of the particles, is largely compensated for by the existence
of an ordered structure-the crystal lattice. The difficulty of deal¬
ing with gases owing to the disordered position of the separate
particles is compensated for by a practically complete absence
of particle interaction. In the case of liquids, however, there
are both kinds of difficulties mentioned above with no corre¬
sponding compensating factors. It can be said that in a liquid
the molecules, as a rule, completely retain their individuality. A
great diversity of motions exists in liquids: displacement of the
molecules, their rotation, vibration of the atoms in the molecules
and vibration of the molecules in the fields of neighbouring
molecules. The worst thing is that all of these types of motion
cannot, strictly speaking, be treated separately (or, as they say, in
the pure form) because there is a strong mutual influence of the

motions.

178 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: I can’t understand how translational motion of
the molecule can be combined with its vibration in the fields of
neighbouring molecules.

TEACHER: Various models have been devised in which attempts
were made to combine these motions. In one model, for in¬
stance, it was assumed that the molecule behaves as follows:
it vibrates for a certain length of time in the field set up by its
neighbours, then it takes a jump, passing over into new sur¬
roundings, vibrates in these surroundings, takes another jump,
etc. Such a model is called the “jump-diffusion model”.

STUDENT B: It seems that is precisely the way in which atoms
diffuse in crystals.

TEACHER: You are right. Only remember that in crystals this
process is slower: jumps into a new environment occur consider¬
ably more rarely. There exists another model according to which
a molecule in a liquid behaves as follows: it vibrates surrounded
by its neighbours and the whole environment smoothly travels
(“floats”) in space and is gradually deformed. This is called the
“continuous-diffusion model”.

STUDENT B: You said that a liquid occupies an intermediate
position between crystals and gases. Which of them is it closer
to?

TEACHER: What do you think?

STUDENT B: It seems to me that a liquid is closer to a gas.

TEACHER: In actuality, however, a liquid is most likely closer to
a crystal. This is indicated by the similarity of their densities,
specific heats and coefficients of volume expansion. It is also
known that the heat of fusion is considerably less than the heat
of vaporization. All these facts are evidence of the appreciable
similarity between the forces of inter-particle bonding in crystals
and in liquids. Another consequence of this similarity is the
existence of elements of ordered arrangement in the atoms of a
liquid. This phenomenon, known as “short-range order”, was
established in A-ray scattering experiments.

WHAT DO YOU KNOW ABOUT THE MOLECULAR-KINETIC THEORY OF MATTER? 179

STUDENT B: What do you mean by short-range order?

TEACHER: Short-range order is the ordered arrangement of a
certain number of the nearest neighbours about any arbitrarily
chosen atom (or molecule). In contrast to a crystal, this ordered
arrangement with respect to the chosen atom is disturbed as
we move away’ from it, and does not lead to the formation of a
crystal lattice. At short distances, however, it is quite similar to
the arrangement of the atoms of the given substance in the solid
ph ase. Shown in Figure 70 (a) is the long-range order for a chain
of atoms. It can be compared with the short-range order shown
in Figure 70 (b).

(a)

0000 0 000 ©

<b)

O000 0 00QQ

The similarity between liquids and crystals has led to the term
“quasi-crystallinity” of liquids.

STUDENT B: But in such a case, liquids can evidently be dealt
with by analogy with crystals.

TEACHER: I should warn you against misuse of the concept of
quasi-crystallinity of liquids and attributing too much impor¬
tance to it. Firstly, you must keep in mind that the liquid state
corresponds to a wide range of temperatures, and the structural-
dynamic properties of liquids cannot be expected to be the same
(or even approximately the same) throughout this range. Near
the critical state, a liquid should evidently lose all similarity to
a solid and gradually transform to the gaseous phase. Thus, the
concept of quasi-crystallinity of liquids may only be justified
somewhere near the melting point, if at all. Secondly, the na¬
ture of the intermolecular interaction differs from one liquid to
another. Consequently, the concept of quasi-crystallinity is not
equally applicable to all liquids. For example, water is found to

Figure 70: Long range order of
crystals and short range order
of liquids.

180 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

be a more quasi-crystalline liquid than molten metals, and this
explains many of its special properties (see § 19).

STUDENT B: I see now that there is no simple picture of the
thermal motions of molecules in a liquid.

TEACHER: You are absolutely right. Only the extreme cases are
comparatively simple. Intermediate cases are always complex.

STUDENT A: The physics entrance examination requirements
include the question about the basis for the molecular-kinetic
theory of matter. Evidently, one should talk about Brownian
motion.

TEACHER: Yes, Brownian motion is striking experimental evi¬
dence substantiating the basic principles of the molecular-kinetic
theory. But, do you know what Brownian motion actually is?

STUDENT A: It is thermal motion of molecules.

TEACHER: You are mistaken; Brownian motion can be observed
with ordinary microscopes! It is motion of separate particles of
matter bombarded by molecules of the medium in their thermal
motion. From the molecular point of view these particles are
macroscopic bodies. Nevertheless, by ordinary standards they
are extremely small. As a result of their random uncompensated
collisions with molecules, the Brownian particles move con¬
tinuously in a haphazard fashion and thus move about in the
medium, which is usually some kind of liquid.

STUDENT B: But why must the Brownian particles be so small?
Why don’t we observe Brownian motion with appreciable parti¬
cles of matter such as tea leaves in a glass of tea?

TEACHER: There are two reasons for this. In the first place, the
number of collisions of molecules with the surface of a particle is
proportional to the area of the surface; the mass of the particle is
proportional to its volume. Thus, with an increase in the size R
of a particle, the number of collisions of molecules with its sur¬
face increases proportionally to R 1 , while the mass of the particle

WHAT DO YOU KNOW ABOUT THE MOLECULAR-KINETIC THEORY OF MATTER? 181

Figure 71: Effect of surface and
volume relationships.

which is to be displaced by the collision increases in propor¬
tion to R 3 . Therefore, as the particles increase in size it becomes
more and more difficult for the molecules to push them about.
To make this clear, I plotted two curves in Figure 71: y = R 2
and y = R' . You can readily see that the quadratic relationship
predominates at small values of R and the cubic relationship at
large values. This means that surface effects predominate at small
values of R and volume effects at large values.

In the second place, the Brownian particle must be very small
since its collisions with molecules are uncompensated, i.e. the
number of collisions from the left and from the right in unit
time should differ substantially. But the ratio of this difference in
the number of collisions to the whole number of collisions will
be the greater, the less the surface of the particle.

STUDENT A: What other facts substantiating the molecular-
kinetic theory are we expected to know?

TEACHER: The very best substantiation of the molecular- kinetic
theory is its successful application in explaining a great number
of physical phenomena. For example, we can give the explana¬
tion of the pressure of a gas on the walls of a vessel containing it.
The pressure p is the normal component of the force F acting on

182 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

unit area of the walls. Since

A (mv)
At

( 100 )

to find the pressure we must determine the momentum trans¬
mitted to a unit area of the wall surface per unit time due to the
blows with which the molecules of the gas strike the walls.

Assume that a molecule of mass m is travelling perpendicular
to a wall with a velocity v. As a result of an elastic collision
with the wall, the molecule reverses its direction of travel and
flies away from the wall with a velocity of v. The change in the
momentum of the molecule equals A (mv) = mAv = 2 mv. This
momentum is transmitted to the wall. For the sake of simplicity
we shall assume that all the molecules of the gas have the same
velocity v and six directions of motion in both directions along
three coordinate axes (assume that the wall is perpendicular to
one of these axes). Next, we shall take into account that in unit
time only those molecules will reach the wall which are at a
distance within v from it and whose velocity is directed toward
the wall. Since a unit volume of the gas contains N/V molecules.

1

in unit time

v molecules strike a unit area of the wall

surface. Since each of these molecules transmits a momentum
of 2 mv, as a result of these blows a unit area of the wall surface

receives a momentum equal to 2mv-

v. According to

equation (100). this is the required pressure p. Thus

2 N mv 2

3 V ~2

( 101 )

According to equation (98), we can replace the energy of the
YYl'V 2 3

molecule by the quantity -kT [in reference to the transla¬
tional motion of molecules, equation (98) is valid for molecules
with any number of atoms]. After this, equation (101) can be
rewritten as

pV = NkT (102)

Note that this result was obtained by appreciable simplification
of the problem (it was assumed, for instance, that the molecules

WHAT DO YOU KNOW ABOUT THE MOLECULAR-KINETIC THEORY OF MATTER?

183

of the gas travel with the same velocity). However, theory shows
that this result completely coincides with that obtained in a
rigorous treatment.

Equation (102) is beautifully confirmed by direct measure¬
ments. It is good proof of the correctness of the concepts of the
molecular-kinetic theory which were used for deriving equation
( 102 ).

Now let us discuss the phenomena of the evaporation and boiling
of liquids on the basis of molecular-kinetic conceptions. How do
you explain the phenomenon of evaporation?

STUDENT A: The fastest molecules of liquid overcome the attrac¬
tion of the other molecules and fly out of the liquid.

TEACHER: What will intensify evaporation?

STUDENT A: Firstly, an increase in the free surface of the liquid,
and secondly, heating of the liquid.

TEACHER: It should be remembered that evaporation is a two-
way process: while part of the molecules leave the liquid, another
part returns to it. Evaporation will be the more effective the
greater the ratio of the outgoing molecules to the incoming
ones. The heating of the liquid and an increase of its free surface
intensify the escape of molecules from the liquid. At the same
time, measures can be taken to reduce the return of molecules
to the liquid. For example, if a wind blows across the surface of
the liquid, the newly escaped molecules are carried away, thereby
reducing the probability of their return. That is why wet clothes
dry more rapidly in the wind.

If the escape of molecules from a liquid and their return compen¬
sate each other, a state of dynamic equilibrium sets in, and the
vapour above the liquid becomes saturated. In some cases it is
useful to retard the evaporation process. For instance, rapid evap¬
oration of the moisture in bread is undesirable. To prevent fast
drying of bread it is kept in a closed container (bread box, plastic
bag). This impedes the escape of the evaporated molecules, and

184 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

a layer of saturated vapour is formed above the surface of the
bread, preventing further evaporation of water from the bread.

Now, please explain the boiling process.

STUDENT A: The boiling process is the same as evaporation, but
proceeds more intensively.

TEACHER: I don’t like your definition of the boiling process at
all. I should mention that many examinees do not understand
the essence of this process. When a liquid is heated, the solubil¬
ity of the gases it contains reduces. As a result, bubbles of gas
are formed in the liquid (on the bottom and walls of the vessel).
Evaporation occurs in these bubbles, they become filled with
saturated vapour, whose pressure increases with the temperature
of the liquid. At a certain temperature, the pressure of the sat¬
urated vapour inside the bubbles becomes equal to the pressure
exerted on the bubbles from the outside (this pressure is equal to
the atmospheric pressure plus the pressure of the layer of water
above the bubble). Beginning with this instant, the bubbles rise
rapidly to the surface and the liquid boils. As you can see, the
boiling of a liquid differs essentially from evaporation. Note that
evaporation takes place at any temperature, while boiling occurs
at a definite temperature called the boiling point. Let me remind
you that if the boiling process has begun, the temperature of the
liquid cannot be raised, no matter how long we continue to heat
it. The temperature remains at the boiling point until all of the
liquid has boiled away.

It is evident from the above discussion that the boiling point
of a liquid is depressed when the outside pressure reduces. In
this, connection, let us consider the following problem. A flask
contains a small amount of water at room temperature. We
begin to pump out the air above the water from the flask with a
vacuum pump. What will happen to the water?

STUDENT A: As the air is depleted, the pressure in the flask will
reduce and the boiling point will be depressed. When it comes
down to room temperature, the water will begin to boil.

WHAT DO YOU KNOW ABOUT THE MOLECULAR-KINETIC THEORY OF MATTER?

185

TEACHER: Could the water freeze instead of boiling?

STUDENT A: I don’t know. I think it couldn’t.

TEACHER: It all depends upon the rate at which the air is pumped
out of the flask. If this process is sufficiently slow, the water
should begin to boil sooner or later. But if the air is exhausted
very rapidly, the water should, on the contrary, freeze. As a re¬
sult of the depletion of the air (and, with it, of the water vapour),
the evaporation process is’ intensified. Since in evaporation the
molecules with the higher energies escape from the water, the
remaining water will be cooled. If the air is exhausted slowly, the
cooling effect is compensated for by the transfer of heat from the
outside. As a result the temperature of the water remains con¬
stant. If the air is exhausted very rapidly, the cooling of the water
cannot be compensated by an influx of heat from the outside,
and the temperature of the water begins to drop. As soon as this
happens, the possibility of boiling is also reduced. Continued
rapid exhaustion of the air from the flask will lower the temper¬
ature of the water to the freezing point, and the unevaporated
remainder of the water will be transformed into ice.

§ 19 How Do You Account For The Peculiarity In
The Thermal Expansion Of Water?

TEACHER: What are the peculiarities of the thermal expansion of
water?

TEACHER: When water is heated from 0 °C to 4 °C its density
increases. It begins to expand only when its temperature is raised
above 4 °C.

TEACHER: How do you explain this?

STUDENT A: I don’t know.

TEACHER: This distinctive feature of water is associated with its
atomic structure. Molecules of water can interact only in one
way: each molecule of water can add on only four neighbouring
molecules whose centres then form a tetrahedron (Figure 72).
This results in a friable, lace-like structure indicative of the quasi¬
crystallinity of water. Of course, we can speak of the structure
of water, as of any other liquid, only on a short-range level (see
§ 18). With an increase in the distance from a selected molecule
this order will undergo gradual distortion due to the bending and
rupture of intermolecular bonds. As the temperature is raised,
the bonds between the molecules are ruptured more frequently,
there are more and more molecules with unoccupied bonds fill¬
ing the vacancies of the tetrahedral structure and, consequently,
the degree of quasi-crystallinity is reduced.

0

Figure 72: Explaining pecu¬
liarity of thermal expansion of
water.

188 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

The above-mentioned lace-like structure of water as a quasi¬
crystalline substance convincingly explains the anomaly of the
physical properties of water, in particular, the peculiarity of its
thermal expansion. On one hand, an increase in temperature
leads to an increase in the mean distances between the atoms and
molecule due to the intensification of intramolecular vibrations,
i. e. the molecules seem to “swell” slightly. On the her hand, an
increase in temperature breaks up the lace-like structure of water
which, naturally, leads to a more dense packing of the molecules
themselves. The first (vibrational) effect should lead to a reduc¬
tion in the density of water. This is the common effect causing
the thermal expansion of solids. The second effect, that of struc¬
ture breakup, should, on the contrary, increase the density of
water as it is heated. In heating water to 4 °C, the structural
effect predominates and the density of water consequently in¬
creases. Upon further heating, the vibrational effect begins to
predominate and therefore the density of water is reduced.

§20 How Well Do You Know The Gas Laws?

TEACHER:

STUDENT

Please write the equation for the combined gas law.
A: This equation is of the form
pV p 0 V 0

T

T n

(103)

where p, V and T are the pressure, volume and temperature of
a certain mass of gas in a certain state, and p 0 , V 0 and T 0 are the
same for the initial state. The temperature is expressed in the
absolute scale.

STUDENT B: I prefer to use an equation of a different form

YYL

pV = —RT (104)

p

where m is the mass of the gas, p is the mass of one gram-
molecule and R is the universal gas constant.

TEACHER: Both versions of the combined gas law are correct.

(To STUDENT B) You have used the universal gas constant. Tell
me, how would you compute its value? I don’t think one can
memorize it.

STUDENT B: To compute R, I can use equation (103), in which
the parameters p 0 , V 0 and T 0 refer to a given mass of gas but
taken at standard conditions. This means that p 0 = 76cm Hg

( YYl \

— ) x 22.41,

p )

190 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

since a gram-molecule of any gas at standard conditions occupies
a definite volume equal to 22.41. The ratio m//u is evidently the
number of gram-molecules contained in the given mass of the
gas. Substituting these values in equation (103) we obtain

76cm Hg x 22.41
273 K

Comparing this with expression (104) we find that R

6.2 (cm Hg)

litres
deg '

TEACHER: I purposely asked you to do these calculations in
order to demonstrate the equivalence of expressions (103) and
(104). Unfortunately, examinees usually know only equation
(103) and are unfamiliar with (104), which coincides with equa¬
tion (102) obtained previously on the basis of molecular-kinetic
considerations. From a comparison of equations (102) and (104)
it follows that ( m/jj)R = Nk. Then

N

R = - ;— -k=N A k (105)

( m\

J

Thus the universal gas constant turns out to be the product of
Avogadro’s number by Boltzmann’s constant. Next, we shall
see whether you can use the equation of the combined gas law.
Please draw a curve showing an isobaric process, i.e. a process in
which the gas pressure remains constant, using coordinate axes V
and T.

STUDENT A: I seem to recall that this process is described by a
straight line.

TEACHER: Why recall? Make use of equation (104). On its basis,
express the volume of the gas as a function of its temperature.

STUDENT A: From equation (104) we get

v= (-)(-) r

(106)

TEACHER: Does the pressure here depend upon the temperature?

HOW WELL DO YOU KNOW THE GAS LAWS?

191

STUDENT A: In the given case it doesn’t because we are dealing
with an isobaric process.

TEACHER:

Good. Then the product

in equation

(106) is a constant factor. We thus obtain a linear dependence of
the volume of the gas on its temperature. Examinees can usually
depict isobaric (p=const), isothermal (T =const) and isochoric
(V=const) processes in diagrams with coordinate axes p and
V. At the same time they usually find it difficult to depict these
processes with other sets of coordinate axes, for instance V and
T or T and p. These three processes are shown in Figure 73 in
different sets of coordinate axes.

T

0

STUDENT B: I have a question concerning isobars in a diagram
with coordinate axes V and T. From equation (106) and from
the corresponding curve in Figure 73 we see that as the tempera¬
ture approaches zero, the volume of the gas also approaches zero.
However, in no case can the volume of a gas become less than
the total volume of all its molecules. Where is the error in my
reasoning?

TEACHER: Equations (102), (103), (104) and (106) refer to the
so-called ideal gas. The ideal gas is a simplified model of a real
gas in which neither the size of the molecules nor their mutual
attraction is taken into consideration. All the curves in Figure 73
apply to such a simplified model, i.e. the ideal gas.

Isotherm

P

Figure 73: The Gas Laws.

Isobar

192 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: But the gas laws agree well with experimental data,
and in experiments we deal with real gases whose molecules have
sizes of their own.

TEACHER: Note that such experiments are never conducted at
extremely low temperatures. If a real gas has not been excessively
cooled or compressed, it can be described quite accurately by the
ideal gas model. Note also that for the gases contained in the air
(for instance, nitrogen and oxygen), these conditions are met at
room temperatures and ordinary pressures.

STUDENT B: Do you mean that if we plot the dependence of the
volume on the temperature in an isobaric process for a real gas,
the curve will coincide with the corresponding straight line in
Figure 73 at sufficiently high temperatures but will not coincide
in the low temperature zone?

TEACHER: Exactly. Moreover, remember that on a sufficiently
large drop in temperature a gas will be condensed into a liquid.

STUDENT B: I see. The fact that the curve of equation (106)
in Figure 73 passes through the origin, or zero point, has no
physical meaning. But then maybe we should terminate the
curve before it reaches this point?

TEACHER: That is not necessary. You are just drawing the curves
for the model of a gas. Where this model can be applied is an¬
other question. Now I want to propose the following. Two
isobars are shown in Figure 74 in coordinate axes V and T : one
corresponds to the pressure p x and the other to the pressure p 2 .
Which of these pressures is higher?

STUDENT A: Most likely, p 2 is higher than p^.

TEACHER: You answer without thinking. Evidently, you decided
that since that isobar is steeper, the corresponding pressure is
higher. This, however, is entirely wrong. The tangent of the

according

to equation (106). It follows that the higher the pressure, the less
the angle of inclination of the isobar. Thus, in our case, p 2 <

Pi

o

Figure 74: Which pressure is
higher?

HOW WELL DO YOU KNOW THE GAS LAWS? 193

p 1 We can reach the same conclusion by different reasoning.

Let us draw an isotherm in Figure 74 (see the dashed line). It
intersects isobar p 2 at a higher value of the gas volume than
isobar p 1 . We know that at the same temperature, the pressure
of the gas will be the higher, the smaller its volume. This follows
directly from the combined gas law [see equation (103) or (104)].
Consequently, p 2 < p\-

STUDENT A: Now, I’m sure I understand.

TEACHER: Then look at Figure 75 which shows two isotherms
(the coordinate axes are p and V) plotted for the same mass of
gas at different temperatures, 7] and T 2 Which is the higher
temperature?

STUDENT A: First I shall draw an isobar (see the dashed line (_ _ )
in Figure 75). At a constant pressure, the higher the temperature
of a gas, the larger its volume. Therefore, the outermost isotherm
T 2 corresponds to the higher temperature.

TEACHER: Correct. Remember: the closer an isotherm is to the
origin of the coordinates p and V, the lower the temperature is.

STUDENT B: In secondary school our study of the gas laws was
of much narrower scope than our present discussion. The com¬
bined gas law was just barely mentioned. Our study was re¬
stricted to Boyle and Mariette’s, Gay-Lussac’s and Charles’ laws.

TEACHER: In this connection, I wish to make some remarks
that will enable the laws of Boyle and Mariotte, Gay-Lussac and
Charles to be included in the general scheme. Boyle and Mari-
otte’s law (more commonly known as Boyle’s law) describes the
dependence of p on V in an isothermal process. The equation
for this law is of the form

higher?

constant

(107)

where the constant

RT.

194 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Gay-Lussac s law describes the dependence of p on T in an
isochoric process. The equation of this law is

p = constant T (108)

where the constant = ( — ] —.

The law of Charles describes the dependence of V on T in an
isobaric process. Its equation is

V = constant T (109)

/ ykl \ R

where the constant = — —. [Equation (109) evidently

V H / P

repeats equation (106).] I will make the following remarks con¬
cerning the above-mentioned gas laws:

(1) All these laws refer to the ideal gas and are applicable to

a real gas only to the extent that the latter is described by
the model of the ideal gas.

(2) Each of these laws establishes a relationship between some
pair of parameters of a gas under the assumption that the
third parameter is constant.

(3) As can readily be seen, each of these laws is a corollary
of the combined gas law [see equation (104)] which es¬
tablishes a relationship between all three parameters
regardless of any special conditions.

(4) The constants in each of these laws can be expressed, not
in terms of the mass of the gas and the constant third pa¬
rameter, but in terms of the same pair of parameters taken
for a different state of the same mass of the gas. In other
words, the gas laws can be rewritten in the following form

(107a)

(108a)

(109a)

HOW WELL DO YOU KNOW THE GAS LAWS? 195

STUDENT A: It seems I have finally understood the essence of the
gas laws.

TEACHER: In that case, let us go on. Consider the following
example. A gas expands in such a manner that its pressure and
volume comply with the condition

pV 2 = constant (110)

We are to find out whether the gas is heated or, on the contrary,
cooled in such an expansion.

STUDENT A: Why must the temperature of the gas change?

TEACHER: If the temperature remained constant, that would

mean that the gas expands according to the law of Boyle and

1

Mariotte [equation (107)]. For an isothermal process poc—,
while in our case the dependence of p on V is of a different
nature: pcc^^).

STUDENT A: Maybe I can try to plot these relationships? The
curves will be of the shape shown in Figure 76.

TEACHER: That’s a good idea. What do the curves suggest?

STUDENT A: I seem to understand now. We can see that in trac-

1

ing the curve poc-^ toward greater volumes, the gas will gradu¬
ally pass over to isotherms that are closer and closer to the origin,
i.e, isotherms corresponding to ever-decreasing temperatures.
This means that in this expansion process the gas is cooled.

TEACHER: Quite correct. Only I would reword your answer. It
is better to say that such a gas expansion process is possible only
provided the gas is cooled.

STUDENT B: Can we reach the same conclusion analytically?

Figure 76: Comparing the

isochores for pcc —!- and
F V 2

1

pcc — .

F V

TEACHER: Of course. Let us consider two states of the gas:
pi, Vj, Tj and p 2 , V 2 , T 2 . Next we shall write the combined

196 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

gas law [see equation (104)] for each of these states

m

P\V\ = RT\

P

p 2 V 2 = —RT 2

p

We can write the given gas expansion process, according to the
condition, in the form

PxVt = p 2 Vi

Substituting the two preceding equations of the gas law in the
last equation, we obtain

-RT l V l = —RT 2 V 2
P (X

After cancelling the common factors we find that

T x V x = T 2 V 2 (111)

From this equation it is evident that if the gas volume is, for
example, doubled, its temperature (in the absolute scale) should
be reduced by one half.

STUDENT A: Does this mean that whatever the process, the gas
parameters (p, V and T) will be related to one another in each
instant by the combined gas law?

TEACHER: Exactly. The combined gas law establishes a relation¬
ship between the gas parameters regardless of any conditions
whatsoever. Now let us consider the nature of the energy ex¬
change between a gas and its environment in various processes.
Assume that the gas is expanding. It will move back all bodies
restricting its volume (for instance, a piston in a cylinder). Con¬
sequently, the gas performs work on these bodies. This work is
not difficult to calculate for isobaric expansion of the gas. As¬
sume that the gas expands isobarically and pushes back a piston
of cross-sectional area S over a distance A/ (Figure 77). The pres¬
sure exerted by the gas on the piston is p. Find the amount of
work done by the gas in moving the piston:

A = FAl = ( pS)Al = p(SAl) = p{V 2 -Vi) (112)

&L

Figure 77: Work done by a gas.

HOW WELL DO YOU KNOW THE GAS LAWS? 197

where V l and V 2 are the initial and final volumes of the gas.

The amount of work done by the gas in nonisobaric expansion
is more difficult to calculate because the pressure varies in the
course of gas expansion. In the general case, the work done
by the gas when its volume increases from V l to V 9 is equal
to the area under the p( V) curve between the ordinates V 1
and V 2 - The amounts of work done by a gas in isobaric and in
isothermal expansion from volume V\ to volume V, are shown
in Figure 78 by the whole hatched area and the crosshatched area,
respectively. The initial state of the gas is the same in both cases.
Thus, in expanding, a gas does work on the surrounding bodies
at the expense of part of its internal energy. The work done by
the gas depends upon the nature of the expansion process. Note
also that if a gas is compressed, then work is done on the gas and,
consequently, its internal energy increases.

The performance of work, however, is not the only method of
energy exchange between a gas and the medium. For example,
in isothermal expansion a gas does a certain amount of work A
and, therefore, loses an amount of energy equal to A. On the
other hand, however, as follows from the principles enumerated
in § 18 [see equation (98)], a constant temperature of the gas in
an isothermal process should mean that its internal energy U
remains unchanged (let me remind you that U is determined by
the thermal motion of the molecules and that the mean energy
of the molecules is proportional to the temperature T). The
question is: what kind of energy is used to perform the work in
the given case?

STUDENT B: Evidently, the heat transmitted to the gas from the
outside.

TEACHER: Correct. In this manner, we reach the conclusion
that a gas exchanges energy with the medium through at least
two channels: by doing work associated with a change in the
volume of the gas, and by heat transfer. The energy balance can
be expressed in the following form

AU = Q-A

(113)

198 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

where A U is the increment of internal energy of the gas charac¬
terized by an increase in its temperature, Q is the heat transferred
to the gas from the surrounding medium, and A is the work done
by the gas on the surrounding bodies. Equation (113) is called
the first law of thermodynamics. Note that it is universal and is
applicable, not only to gases, but to any other bodies as well.

STUDENT B: To sum up, we may conclude that in isothermal
expansion, all the heat transferred to the gas is immediately
converted into work done by the gas. If so, then isothermal
processes cannot take place in a thermally insulated system.

TEACHER: Quite true. Now consider isobaric expansion of gas
from the energy point of view.

STUDENT B: The gas expands. That means that it performs
work. Here, as can be seen from equation (106), the temperature
of the gas is raised, i.e. its internal energy is increased. Conse¬
quently, in this case, a relatively large amount of heat must be
transferred to the gas: a part of this heat is used to increase the
internal energy of the gas and the rest is converted into the work
done by the gas.

TEACHER: Very good. Consider one more example. A gas is
heated so that its temperature is increased by AT. This is done
twice: once at constant volume of the gas and then at constant
pressure. Do we have to expend the same amount of heat to heat
the gas in both cases?

STUDENT A: I think so.

STUDENT B: I would say that different amounts are required. At
constant volume, no work is done, and all the heat is expended
to increase the internal energy of the gas, i.e. to raise its tempera¬
ture. In this case

Qi = CjAT (114)

At constant pressure, the heating of the gas is inevitably asso¬
ciated with its expansion, so that the amount of work done is
A = p{V — V]). The supplied heat Q 2 is used partly to increase

HOW WELL DO YOU KNOW THE GAS LAWS?

199

the internal energy of the gas (to raise its temperature) and partly
to do this work. Thus

Q 2 = C 1 AT + p(V-V l ) (115)

Obviously, Qj < q 2 -

TEACHER: I agree with Student B. What do you call the quan¬
tity of heat required to raise the temperature of a body by one
degree?

STUDENT B: The heat capacity of the body.

TEACHER: What conclusion can be drawn from the last example
as regards the heat capacity of a gas?

STUDENT B: A gas has two different heat capacities: at constant
volume and at constant pressure. The heat capacity at constant
volume (which is factor Cj in the last two equations) is less than
the heat capacity at constant pressure.

TEACHER: Can you express the heat capacity at constant pressure
in terms of Q, that is, the heat capacity at constant volume?

STUDENT B: I’ll try. Let us denote the heat capacity at constant
pressure by C 2 In accordance with the definition of heat capacity,
we can write C 2 = Q-,/AT. Substituting the value of Q 9 from
equation (115) we obtain

C 2 = C, +

P{V-V i)

AT

(116)

TEACHER: You stopped too soon. If we apply the equation of the
combined gas law, we can write

m yyi

p{V - Vi) = — R(T - 7i) = —RAT
T T

after substituting into equation (116), we obtain

_ _ m

c 2 = q + —r

( 117 )

200 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

In reference to one gram-molecule of the gas ( m. = pi), this
relationship is even more simple:

C 2 = C 1 +R

(118)

In conclusion, let us consider a certain cycle consisting of an
isotherm, isochore and isobar (see Figure 79 (a) in which axes
p and V are used as the coordinate axes). Please draw this same
cycle (qualitatively) in a diagram with coordinate axes V and T,
and analyse the nature of energy exchange between the gas and
the medium in each element of the cycle.

STUDENT B: In a diagram with coordinate axes V and T, the
cycle will be of the form illustrated in Figure 79 (b).

TEACHER: Quite correct. Now please analyse the nature of
energy exchange between the gas and the medium in the separate
elements of the cycle.

STUDENT B: In element 1-2, the gas undergoes isothermal expan¬
sion. It receives a certain amount of heat from the outside and
spends all this heat in doing work. The internal energy of the gas
remains unchanged.

In element 2-3 of the cycle, the gas is heated isochorically (at
constant volume). Since its volume does not change, no work is
done. The internal energy of the gas is increased only due to the
heat transferred to the gas from the outside.

In element 3-1 the gas is compressed isobarically (at constant
pressure) and its temperature drops as can be seen in Figure 79 (b).
Work is done on the gas, but its internal energy is reduced. This
means that the gas intensively gives up heat to the medium.

TEACHER: Your reasoning is absolutely correct.

STUDENT A: Our discussion shows me that I knew very little
about the gas laws. Do we have, to know all this for the entrance
examinations? In my opinion, some of the questions we dis¬
cussed are beyond the scope of the physics syllabus for students
taking entrance examinations.

(O)

HOW WELL DO YOU KNOW THE GAS LAWS? 201

TEACHER: If you carefully think over our discussion, you will see
that it only covered questions directly concerned with the com¬
bined gas law in its general form or as applied in certain special
cases. Your confusion should be attributed not to the imagi¬
nary stretching of the syllabus, but simply to the fact that you
have not thought over and understood the gas laws thoroughly
enough. Unfortunately, examinees frequently don’t care to go
beyond a very superficial idea of the gas laws.

§27 How Do You Go About Solving Problems
On Gas Laws?

STUDENT A: I would like to look into the application of gas laws
in solving various types of problems.

TEACHER: In my opinion, almost all the problems involving gas
laws that are assigned to examinees are quite simple. Most of
them belong to one of the following two groups.

First group: Problems devised on the basis of a change in the
state in a certain mass of gas; the value of the mass is
not used. As a result of expansion, heating and other
processes, the gas goes over from a certain state with pa¬
rameters pi, V 1 and Tj to a state with parameters p 2 , V 2
and T 2 . The parameters of the initial and final states are
related to one another by the equation of the combined
gas law

P\V\ Pl^2

(119)

The problem consists in finding one of these six parame¬
ters.

Second group: Problems in which the state of the gas does not
change but the value of the mass of the gas appears in the
problem. It is required to find either this mass when all
the parameters are known, or one of the parameters when
the mass and the other parameters are known. In such
problems the molecular weight of the gas must be known.

204 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: I think the most convenient way of solving prob¬
lems of the second group is to use equation (104) of the com¬
bined gas law.

TEACHER: Of course you can use this equation. To do this, how¬
ever, you must know the numerical value of the universal gas
constant R. Asa rule, nobody remembers it. For this reason, in
practice it is more convenient to resort to the following method:
we assume that the gas is brought to standard conditions, de¬
noting the gas parameters at these conditions by p s , V s and T s ;
Then we can write the equation

pV _ p s V s

( 120 )

where

V s = — x 22.41

STUDENT B: In my opinion, this method of solution is by no
means simpler than the use of equation (104). Here we have
to remember three values: P 5 = 76cm Hg, T s = 273 K and
V$/{m/pi ) = 22.41. It is obviously simpler to memorize one
value, the universal gas constant.

TEACHER: Nevertheless, my method is simpler because nobody
has any difficulty in remembering the three values you indicated
(pressure, temperature and the volume of the gram-molecule of
a gas under standard conditions). Assume that we are to find the
volume of 58 g of air at a pressure of 8 atm and a temperature
of 91 °C. Let us solve this problem by the method I proposed.
Since the mass of the gram-molecule of air equals 29 g, we have 2
gram-molecules. At standard conditions they occupy a volume of
44.81. From equation (120) we obtain

1 x 364

-=7.51

8x273

STUDENT B: I see you have assumed that p s = 1 atm. The condi¬
tions of the problem, however, most likely referred to technical
atmospheres. Then it should be p s = 1.034 atm.

HOW DO YOU GO ABOUT SOLVING PROBLEMS ON GAS LAWS? 205

TEACHER: You are right. There is a difference between the phys¬
ical atmosphere (corresponding to standard pressure) and the
technical atmosphere. I simply neglected this difference.

STUDENT A: Could you point out typical difficulties in solving
problems of the first and second groups?

TEACHER: I have already mentioned that in my opinion these
problems are quite simple.

STUDENT A: But what mistakes do examinees usually make?

TEACHER: Apart from carelessness, the main cause of errors is
the inability to compute the pressure of the gas in some state or
other. Consider a problem involving a glass tube sealed at one
end. The tube contains a column of mercury isolating a certain
volume of air from the medium. The tube can be turned in a
vertical plane. In the first position (Figure 80(a)), the column
of air in the tube has the length /j and in the second position
(Figure 80(b)), / 2 . Find the length / 3 of the column of air in
the third position when the tube is inclined at an angle of a to
the vertical (Figure 80(c)). We shall denote the atmospheric
pressure by p A in terms of length of mercury column, and the
length of the mercury column in the tube by A/. In the first
position, the pressure of the air in the tube is evidently equal
to the atmospheric pressure. In the second position, it is equal
to the difference (p A — A/), because the atmospheric pressure
is counterbalanced by the combined pressures of the mercury
column and the air inside the tube. Applying the law of Boyle
and Mariotte we write

(a)

J

MR

- 1

Al t

hpA ~ h(pA~Al)

from which we find that the atmospheric pressure is

Pa = AI-\ ( 121 )

h —

In the third position, a part of the weight of the mercury column
will be counterbalanced by the reaction of the tube walls. As a
result, the pressure of the air inside the tube turns out to be equal

206 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

to (pA — A/ cos a). Using the law of Boyle and Mariotte for the
first and third states of the gas, we can write

1\Va = h (Pa — A/ cos a)

from which the atmospheric pressure equals

Pa

A /

/ 3 cos a

( 122 )

Equating the right-hand sides of equations (121) and (122) we
obtain

l 2 It, cos a

h ~h ^3 h

from which we find the required length

h

_ hh _

l 2 — (/ 2 — I]) cos a

(123)

You can readily see that if cos a = 1, then / 3 = l 2 i. e. we have the
second position of the tube, and if cos a = 0, then / 3 = / 3 which
corresponds to the first position of the tube.

STUDENT A: The first and second groups of problems in your
classification are clear to me. But, is it likely that the examination
will include combinations of problems from the first and second
groups?

TEACHER: Why yes, such a possibility cannot be ruled out. Let
us consider the following problem. At a pressure of 2 atm, 16 g of
oxygen occupies a volume of 5 litres. How will the temperature
of the gas change if it is known that upon increase in pressure to
5 atm the volume reduces by 1 litre?

STUDENT A: The mass, pressure and volume of the oxygen being
known, we can readily find its temperature. Thus, 16 g of oxygen
is 0.5 gram-molecule,which has a volume of 11.2 litres at standard
conditions. Next, we find that

A

r hYl

S PsVs

= 273

2x5
1 x 11.2

= 244 K

(124)

HOW DO YOU GO ABOUT SOLVING PROBLEMS ON GAS LAWS?

207

TEACHER: Quite right. At the given stage you’ve handled the
problem as a typical one from the second group.

STUDENT A: Then, since we know the temperature T 1 of the gas
in the initial state, we can find the temperature T 2 in the final
state. Thus

„ „ p7 V 2 5x4

7, = 7,= 244-= 488 K

2 p x V x 2x5

Comparing this result with equation (124), we find that the
temperature has been raised by 244 K.

TEACHER: Your solution is absolutely correct. As you could see,
the second half of the problem was dealt with as a typical one
from the first group.

STUDENT B: At the very beginning of our discussion, in speaking
of the possible, groups of problems, you said most problems be¬
long to these groups. Are there problems that differ in principle
from those of the first and second groups?

TEACHER: Yes, there are. In the problems of these groups, it was
assumed that the mass of the gas remained unchanged. Problems
can be devised, however, in which the mass of the gas is changed
(gas is pumped out of or into the container). We will arbitrarily
classify such problems in the third group. There are no ready¬
made rules for solving such problems; they require an individual
approach in each case. However, in each specific case, problems
of the third group can be reduced to problems of the first two
groups or to their combination. This can be illustrated by two
examples.

Here is the first one. The gas in a vessel is subject to a pressure of
20 atm at a temperature of 27°C. Find the pressure of the gas in
the vessel after one half of the mass of the gas is released from the
vessel and the temperature of the remainder is raised by 50°.

This problem resembles those of the first group, since it involves
a change in the state of the gas. With the change in state, how¬
ever, the mass of the gas also changes. In order to make use of the

208 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

combined gas law, we must study the change in state of the same
mass of the gas. We shall choose the mass of the gas that is finally
left in the vessel. We denote its final parameters by p 2 , V 2 and
T 2 . Then T 2 = (273 + 27 + 50) = 350 K; V 2 = V, where V is the
volume of the vessel; and is the required pressure. How can
we express the initial parameters of this mass of gas?

STUDENT A: It will have the same temperature as the whole mass
of gas: Tj = (273 + 27) = 300 K; its volume will be one half of the
volume of the vessel, i.e. V /2; and its pressure is the same as that
of the whole mass of gas: p x = 20 atm.

STUDENT B: I would deal with the initial parameters of the
above-mentioned mass of gas somewhat differently: 7) = 300 K:
the volume is the same as for the whole mass of gas (V) = V) but
the pressure is equal to one half of the pressure of the whole mass
of gas, i.e. p { = 10 atm.

TEACHER: Since the pressure and volume appear in the equation
in the form of their product, both of your proposals, though
they differ, lead to the same result. For this reason, we could
have, refrained from discussing these differences if they didn’t
happen to be of interest from the physical point of view. We
shall arbitrarily call the molecules of the portion of the gas that
finally remains in the vessel “white” molecules, and those of the
portion to be released from the vessel, “black” molecules. Thus,
we have agreed that the white molecules remain in the vessel and
the black molecules are released from it. The initial state of the
gas can be treated in two ways:

(a)

( 0 )

Figure 81: Work done by a gas.

(1) the black and white molecules are separated so that macro¬
scopic volumes can be separated out in the vessel contain¬
ing only white or only black molecules (Figure 81(a));

(2) the white and black molecules are thoroughly mixed
together so that any macroscopic’ volume contains a
practically equal number of each kind of molecules (Fig¬
ure 81(b)).

In the first case, molecules of each kind form their own gaseous

HOW DO YOU GO ABOUT SOLVING PROBLEMS ON GAS LAWS? 209

“body” with a volume of V /2 which exerts a pressure of 20 atm
on the walls and on the imaginary boundary with the other
body. In the second case, molecules of both kinds are distributed
uniformly throughout the whole volume V of the vessel, and the
molecules of each kind exert only one half of the pressure on the
walls (at any place on the walls, one half of the blows come from
white molecules and the other half from black ones). In this case,
Vj = V and p 1 = 10 atm. In connection with this last remark, let
us recall the law of partial pressures: the pressure of a mixture of
gases is equal to the sum of the pressures of the component gases.
I wish to emphasize that here we are dealing with a mixture of
gases, where molecules of all kinds are intimately mixed together.

STUDENT B: I think the second approach is more correct because
the molecules of both kinds are really mixed together.

TEACHER: In the problem being considered, both approaches are
equally justified. Don’t forget that our a priori division of the
molecules into two kinds was entirely arbitrary.

But let us return to the solution of the problem. We write the
equation of the combined gas law for the mass of the gas remain¬
ing in the vessel:

10V p 2 V

W ~~ lio"

from which we find that p 2 = 11.7 atm.

Now, consider the following problem. A gas is in a vessel of
volume V at a pressure of p 0 . It is being pumped out of the
vessel by means of a piston pump with a stroke volume of v
(Figure 82). Find the number, n, of strokes required to lower the
pressure of the gas in the vessel to p n .

STUDENT A: This problem seems to be quite simple: n strokes
of the piston lead to an n-fold increase in the volume of the gas
by the volume v. Therefore, we can write the law of Boyle and
Mariotte in the form

Po V = Pn( V + nv )

Figure 82: Work done by a gas.

from which we can find the number of strokes n.

210 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: To what mass of gas does your equation refer?

STUDENT A: To the mass that was initially in the vessel.

TEACHER: But even after the first stroke a part of this mass
leaves the system entirely: when the piston moves to the left it
closes valve A and opens valve B through which the gas leaves
the system (see Figure 82). In other words, the w-fold increase of
the volume of the gas by the amount v does not refer to the same
mass of gas. Consequently, your equation is incorrect.

Let us consider each stroke of the piston separately. We shall
begin with the first stroke. For the mass of gas that was initially
in the vessel we can write

p 0 v = p 1 (V + v)

where p 1 is the pressure of the gas after the piston has completed
the first working stroke and is in the extreme right-hand posi¬
tion. Then the piston returns to its initial left-hand position. At
this, as I previously mentioned, valve A is closed, and the mass of
the gas in the vessel is less than the initial mass. Its pressure is p 1 .
For this mass of gas we can write the equation

P\V = p 2 ( v + v)

where p 2 is the pressure of the gas at the end of the second
stroke. Dealing consecutively with the third, fourth and sub¬
sequent strokes of the piston, we obtain a system of equations of
the law of Boyle and Mariotte:

Po V = Pi( V + v) '

P\V = p 2 (V +v)

p 2 V = p,(V + v)\ (125)

Pn-lV = P„(V + v))

Each of these equations refers to a definite mass of gas. Solving
the system of equations (125) we obtain

Pn ~ Po

V

V + v

HOW DO YOU GO ABOUT SOLVING PROBLEMS ON GAS LAWS? 211

Taking the logarithm of this result, we finally obtain

n =

(126)

Problems

39. A glass tube with a sealed end is completely submerged in a vessel
with mercury (Figure 83). The column of air inside the tube has a
length of 10 cm. To what height must the upper end of the tube be
raised above the level of the mercury in the vessel so that the level
of the mercury inside the tube is at the level of the mercury in the
vessel? Assume standard atmospheric pressure. Calculate the mass
of the air inside the tube if its cross-sectional area equals 1 cm 2 . The
temperature is 27 °C.

Figure 83: A glass tube with
sealed end submerged in a
vessel with mercury.

40. A glass tube, one end of which is sealed, is submerged with the open
end downward into a vessel with mercury (Figure 84). How will the
level of the mercury in the tube change if the temperature is raised
from 27 °C to 77 °C? Neglect the thermal expansion of the tube.
Assume standard atmospheric pressure. Find the mass of the air
inside the tube if its cross- sectional area is 0.5 cm 2 .

41. The air in a vessel with a volume of 5 L has a temperature of 27 °C
and is subject to a pressure of 20 atm. What mass of air must be
released from the vessel so that its pressure drops to 10 atm?

42. Compute the amount of work done by a gas which is being iso-
barically heated from 20 °C to 100 °C if it is in a vessel closed by a
movable piston with a cross-sectional area of 20 cm 2 and weighing 5
kgf. Consider two cases:

Figure 84: A glass tube with
sealed end submerged in a
vessel with mercury.

(a) the vessel is arranged horizontally (Figure 85 (a)), and

(b) the vessel is arranged vertically (Figure 85 (b)).

The initial volume of the gas is 5 litres. Assume standard atmo¬
spheric pressure.

(a> (b)

4

Figure 85: Work done by a gas.

212 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

43. A column of air 40 cm long in a glass tube with a cross-sectional
area of 0.5 cm 2 and arranged vertically with the sealed end upward,
is isolated by a column of mercury 8 cm long. The temperature is
27 °C. How will the length of the air column change if the tube is
inclined 60° from the vertical and the temperature is simultaneously
raised by 30°? Assume standard atmospheric pressure. Find the
mass of the air enclosed in the tube.

44. What is the mass of the water vapour in a room of a size 6m x 5 m x
3.5m if, at a temperature of 15°C 15 °C, the relative humidity is
55%? Will dew be formed if the air temperature drops to 10 °C?
What part of the total mass of the air in the room is the mass of the
water vapour if the air pressure equals 75 cm Hg?

What is a field? How is a field described? How does motion take place
in a field? These fundamental problems of physics can be most conve¬
niently considered using an electrostatic field as an example. We shall
discuss the motion of charged bodies in a uniform electrostatic field. A
number of problems illustrating Coulomb's law will be solved.

§22 Let Us Discuss Field Theory

TEACHER: Let us discuss the field, one of the basic physical con¬
cepts. For the sake of definiteness, we shall deal with electrostatic
fields. What is your idea of a field? What do you think it is?

STUDENT A: I must confess that I have a very vague idea of what
a field really is. A field is something elusive, invisible, a kind of
spectre. At the same time, it is said to be present throughout
space. I do not object to the field being defined as a material
entity. But this means nothing to me. When we speak of matter,

I understand what we are talking about. But when we speak of a
field, I give up.

STUDENT B: To me, the concept of the field is quite tangible.

Matter in any substance is in concentrated form, as it were. In
a field, on the contrary, matter is “spread” throughout space,
so to speak. The fact that we cannot see a field with the naked
eye doesn’t prove anything. A field can be “seen” excellently by
means of relatively simple instruments. A field acts as a transmit¬
ter of interactions between bodies. For instance, an electrostatic
field transmits interactions between fixed electric charges. Each
charge can be said to set up a field around itself. A field set up by
one charge influences another charge and, conversely, the field
set up by the second charge influences the first charge. Thus,
Coulomb (electrostatic) interaction of charges is accomplished.

STUDENT A: But couldn’t we get along without any “go-betweens”?
What prevents us from supposing that one charge acts directly on
another charge?

216 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: Your supposition may raise serious objections.
Assume that at some instant one of the charges is displaced (i.e.
“budges”) for some reason. If we proceed from the supposition
of, “direct interaction”, we have to conclude that the second
charge must also “budge” at the very same instant. This would
mean that a signal from the first charge reaches the second charge
instantaneously. This would contradict the basic principles of
the theory of relativity. If, however, we have a transmitter of
interactions, l.e. a field, the signal is propagated from one charge
to the other through the field. However large the velocity of
propagation, it is always finite. Therefore, a certain interval of
time exists during which the first charge has stopped “budging”
and the second has not yet started. During this interval, only the
field contains the signal for “budging”.

STUDENT A: All the same, I would like to hear a precise defini¬
tion of the field.

TEACHER: I listened to your dialogue with great interest. I feel
that STUDENT B has displayed a keen interest in problems of
modern physics and has read various popular books on physics.
As a result, he has developed what could be called initiative
thinking. To him the concept of the field is quite a real, “work¬
ing” concept. His remarks on the field as a transmitter of interac¬
tions are quite correct. STUDENT A, evidently, confined himself
to a formal reading of the textbook. As a result, his thinking is
inefficient to a considerable extent. I say this, of course, without
any intention to offend him or anyone else, but only to point
out that many examinees feel quite helpless in like situations.
Strange as it may be, a comparatively large number of students
almost never read any popular-science literature. However, let
us return to the essence of the problem. (To STUDENT A) You
demanded a precise definition of the field. Without such a def¬
inition the concept of the field eludes you.However, you said
that you understand what matter is. But do you really know the
precise definition of matter?

STUDENT A: The concept of matter requires no such definition.
Matter can be “touched” with your hand.

LET US DISCUSS FIELD THEORY

217

TEACHER: In that case, the concept of the field also “requires no
such definition”; it can also be “touched”, though not with your
hand. However, the situation with the definition is much more
serious. To give a precise, logically faultless definition means
to express the concept in terms of some more “primary” con¬
cepts. But what can be done if the given concept happens to be
one of the “primary” concepts? Just try to define a straight line
in geometry. Approximately the same situation exists with re¬
spect to the concepts of matter and the field. These are primary,
fundamental concepts for which we can scarcely hope to find a
clear-cut blanket definition.

STUDENT A: Can we, nevertheless, find some plausible defini¬
tion?

TEACHER: Yes, of course. Only we must bear in mind that no
such definition can be exhaustive. Matter can exist in various
forms. It can be concentrated within a restricted region of space
with more or less definite boundaries (or, as they say, it can be
“localized”), but, conversely, it can also be “delocalized”. The
first of these states of matter can be associated with the concept
of matter in the sense of a “substance”, and the second state
with the concept of the “field”. Along with their distinctive
characteristics, both states have common physical characteristics.
For example, there is the energy of a unit volume of matter
(as a substance) and the energy of a unit volume of a field. We
can speak of the momentum of a unit volume of a substance
and the momentum of a unit volume of a field. Each kind of
field transmits a definite kind of interaction. It is precisely from
this interaction that we can determine the characteristics of the
field at any required point. An electrically charged body, for
instance, sets up an electrostatic field around itself in space. To
reveal this field and measure its intensity at some point in space,
it is necessary to bring another charged body to this point and
to measure the force acting on it. It is assumed that the second
charged body is sufficiently small so that the distortion it causes
in the field can be neglected.

The properties of matter are inexhaustible, the process of seeking

218 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

knowledge is eternal. Gradually, step by step, we advance along
the road of learning and the practical application of the proper¬
ties of matter that surrounds us. In our progress we have to “stick
on labels” from time to time which are a sort of landmarks along
the road to knowledge. Now we label something as a “field”. We
understand that this “something” is actually the primeval abyss.
We know much about this abyss we have called a “field”, and
therefore we can employ this newly introduced concept more
or less satisfactorily. We know much, but far from all. An at¬
tempt to give it a clear-cut definition is the same as an attempt to
measure the depth of a bottomless chasm.

STUDENT B: I think that the concept of the field, as well as any
other concept emerging in the course of our study of the material
world, is inexhaustible. This, exactly, is the reason why it is
impossible to give an exhaustive, clear-cut definition of a field.

TEACHER: I completely agree with you.

STUDENT A: I am quite satisfied with your remarks about sub¬
stance and the field as two states of matter-localized and unlo¬
calized. But why did you begin this discussion about the inex¬
haustibility of physical concepts and the eternity of learning?

As soon as I heard that, clarity vanished again and everything
became sort of blurred and vague.

TEACHER: I understand your state of mind. You are seeking for
some placid definition of a field, even if it is not absolutely pre¬
cise. You are willing to conscientiously memorize this definition
and hand it out upon request. You don’t wish to recognize that
the situation is not at all static but, on the contrary, a dynamic
one. You shouldn’t believe that everything becomes blurred
and vague. I would say that everything becomes dynamic in the
sense that it tends toward change. Any precise definition, in it¬
self, is something rigid and final. But physical concepts should
be investigated in a state of their development. That which we
understood to be the concept of the field yesterday appreciably
differs from what we understand by this concept today. Thus,
for instance, modern physics, in contrast to the classical version,

LET US DISCUSS FIELD THEORY

219

does not draw a distinct boundary between the field and sub¬
stance. In modern physics, the field and substance are mutually
transformable: a substance may become a field and a field may
become a substance. However, to discuss this subject in more
detail now would mean getting too far ahead.

STUDENT B: Our discussion on physics has taken an obviously
philosophical turn.

TEACHER: That is quite natural because any discussion of phys¬
ical conceptions necessarily presupposes that the participants
possess a sufficiently developed ability for dialectical thinking.

If this ability has not yet been cultivated, we have, even against
our will, to resort to digressions of a philosophical nature. This
is exactly why I persist in advising you to read more and more
books of various kinds. Thereby you will train your thinking
apparatus, make it more flexible and dynamic. In this connec¬
tion, invaluable aid can be rendered to any young person by V. I.
Lenin’s book Materialism and Empiriocriticism. I advise you to
read it.

STUDENT A: But that is a very difficult book. It is studied by
students of institutes and universities.

TEACHER: I don’t insist on your studying this book. It certainly
was not intended for light reading. Simply try to read it through
carefully. Depending on your background, this book will exert a
greater or lesser influence on your mode of thinking. In any case,
it will be beneficial.

In conclusion, I wish to mention the following: Student A is
obviously afraid of vagueness or indefiniteness; he demands
maximum precision. He forgets that there is a reasonable limit to
everything, even precision. Try to imagine a completely precise
world about which we have exhaustive information. Just conjure
up such a world and then tell me: wouldn’t you be amazed at
its primitiveness and inability to develop any further? Think
about all this and don’t hurry with your conclusions. And now,
for the present, let us attempt to approach the problem from
another angle. I will pose the following question: “How is a field

220 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

described?” I know that many people, after getting the answer,
will say: “Now we know what the field is”.

§ 23 How Is An Electrostatic Field Described?

TEACHER: Thus, we continue the discussion that we began in
the preceding section by asking: “How is an electrostatic field
described?”

STUDENT B: An electrostatic field is described by means of a
vectorial force characteristic called the intensity of the electric
field. At each point in the field, the intensity E has a definite
direction and numerical value. If we move from one point in
a field to another in such a manner that the directions of the
intensity vectors are always tangent to the direction of motion,
the paths obtained by such motion are called the lines of force
of the field. Lines of force are very convenient for graphically
representing a field.

TEACHER: Good. Now let us reason more concretely. The
Coulomb force of interaction between two charges q x and q 2
spaced a distance of r apart can be written in the form

r _ ^2
e 2

Y L

This can be rewritten

as

E(r) =

<h

(127)

(128)

F e =E(r)q 2 (129)

Equation (128) signifies that charge q l sets up a field around
itself, whose intensity at a distance of r from the charge equals

222 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Equation (129) signifies that this field acts on charge q 0
r 2

located at a distance of r from charge q l with a force E(r)q 2 .
Equation (127) could be written thus because a “go-between” -
the quantity E, the characteristic of the field - was introduced.
Try to determine the range of application of equations (127),

(128) and (129).

STUDENT B: Equation (127) is applicable for two point charges.
That means that the range of application of equations (128) and

(129) is the same. They were obtained from equation (127).

TEACHER: That is correct only with respect to equations (127)
and (128). Equation (129) has a much wider range of application.
No matter what sets up the field E (a point charge, a set of point
charges or of charged bodies of arbitrary shape), in all cases the
force exerted by this field on charge q Q is equal to the product
of this charge by field intensity at the point where charge q 0
is located. The more general version of equation (129) has the
following vectorial form

F e = E( r )q 0 (130)

where the arrows, as usual, serve to denote the vectors. It is evi¬
dent from equation (130) that the direction of the force acting on
charge q 0 at the given point of the field coincides with the direc¬
tion of the field intensity at this point if charge q 0 is positive. If
charge q 0 is negative, the direction of the force is opposite to the
intensity.

Here we can sense the independence of the concept of the field.
Different charged bodies set up different electrostatic fields, but
each of these fields acts on a charge situated in it according to the
same law (130). To find the force acting on a charge, you must
first calculate the intensity of the field at the point where the
charge is located. Therefore, it is important to be able to find the
intensity of the field set up by a system of charges. Assume that
there are two charges, q 1 and q 2 ■ The magnitude and direction
of the intensity of the field set up by each of these charges can
readily be found for any point in space that may interest us.

HOW IS AN ELECTROSTATIC FIELD DESCRIBED? 223

Assume that at a certain point, specified by the vector f these
intensities are described by the vectors E^(r) and E 2 (r). To find
the resultant intensity at point r , you must add vectorially the
intensities due to the separate charges

£(r)=£ 1 (r)+£ 2 (r) (131)

I repeat that the intensities must be added vectorially. (To STU¬
DENT A) Do you understand?

STUDENT A: Yes, I know that intensities are added vectorially.

TEACHER: Good. Then we can check how well you can use this
knowledge in practice. Please draw the lines of force of the field
of two equal and opposite charges ( +q j and — q 2 ) assuming that
one of the charges (for instance, +qi) is several times greater than
the other.

STUDENT A: I’m afraid I can’t. We never discussed such fields
before.

TEACHER: What kind of fields did you study?

STUDENT A: I know what the picture of the lines of force looks
like for a field set up by two point charges of equal magnitude. I
have drawn such a picture in Figure 86.

Figure 86: Lines of force
looks like for a field set up
by two point charges of equal
magnitude.

TEACHER: Your drawing is somewhat inaccurate though quali¬
tatively it does represent the force lines of a field set up by two
charges of the same magnitude but of opposite sign. Can’t you

224 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

vizualise how this picture will change as one of the charges in¬
creases?

STUDENT A: We never did anything like that.

TEACHER: In that case, let us use the rule for the vectorial addi¬
tion of intensities. We shall begin with the familiar case when the
charges are equal (Figure 87 (a)). We select three points A, B and
C and construct a pair of intensity vectors for each point: E l and
E 2 (Ej for the field of charge +q x and E 1 for the field of charge
—q 2 ). Then we add the vectors E^ and E 2 for each of these points
to obtain the resultant vectors E A , E B and E c . These vectors
must be tangent to the lines of force of the field at the corre¬
sponding points. These three vectors indicate the behaviour of
the lines of force which are shown in Figure 88 (a). Compare this
drawing with Figure 86 proposed by you. Note your inaccuracies
in the behaviour of the lines of force to the left of charge — q and
to the right of charge +q. Assume now that charge +q x is dou¬
bled in magnitude, and charge — q 2 is halved (Figure 87 (b)). We
select, as before, three points A, B and C. First we construct the
intensity vectors for these points and then find their resultants:
E A , E B and E c . The picture of the lines of force corresponding
to these vectors is shown in Figure 88 (b).

Figure 87: Lines of force
looks like for a field set up
by two point charges of equal
magnitude.

Finally, we assume that q x is doubled again and that q 2 is halved
again (Figure 87 (c)). Next we construct the resultant vectors E A ,
Eg and E c for points A, B and C. The corresponding picture of

Figure 88: Lines of force
looks like for a field set up
by two point charges of equal
magnitude.

HOW IS AN ELECTROSTATIC FIELD DESCRIBED?

225

the lines of force is shown in Figure 88 (c).

As you see, the influence of charge +q x becomes greater with an
increase in its relative magnitude; the field of charge +q^ begins
to repress the field of charge — q 2 -

STUDENT A: Now I understand how to construct a picture of the
lines of force for a field set up by a system of several charges.

TEACHER: Let us continue our discussion of an electrostatic
field. This field has one important property which puts it in
the same class with gravitational fields, namely: the work done
by the forces of the field along any closed path equals zero. In
other words, if the charge travelling in the field returns to its
initial point of departure, the work done by the forces of the field
during this motion is equal to zero. Over certain portions of the
path this work will be positive and over others negative, but the
sum of the work done will equal zero. Interesting consequences
follow from this property of an electrostatic field. Can you name
them?

STUDENT B: No, I can’t think of any.

TEACHER: I’ll help you. You probably have noted that the lines
of force of an electrostatic field are never closed on themselves.
They begin and end in charges (beginning in positive charges and
terminating in negative ones) or they end at infinity (or arrive
from infinity). Can you associate this circumstance with the
above-mentioned property of the electrostatic field?

Now I seem to understand. If a line of force in an electrostatic
field was closed on itself, then by following it we could return
to the initial point. As a charge moves along a line of force, the
sign of the work done by the field evidently does not change
and, consequently, cannot be equal to zero. On the other hand,
the work done along any closed path must a be equal to zero.
Hence, lines of force of an electrostatic field cannot be closed on
themselves.

TEACHER: Quite correct. There is one more consequence follow¬
ing from the above-mentioned property of the electrostatic field:

226 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

the work done in moving a charge from one point of the field to
another does not depend upon the path followed.

We can move a charge from point a to point b, for instance,
along different paths, 1 and 2 (Figure 89). Let us denote by A l
the amount of work done by the forces of the field to move the
charge along path 1 and that along path 2 by A 2 Let us accom¬
plish a complete circuit: from point a to point b along path 1 and
from point b back to point a along path 2. During the re- turn
along path 2, the work done will be —A 2 . The total work done in
a complete circuit is, A 1 + {—A 2 ) = ^1 ~A 2 - Since the work done
along any path closed on itself equals zero, then A j = A 2 . The
fact that the work done in moving a charge is independent of the
chosen path but depends only on the initial and final points, en¬
ables this value to be used as a characteristic of the field (since it
depends only upon the chosen points of the field!). Thus another
characteristic of an electrostatic field, its potential, is introduced.
In contrast to the intensity, this is a scalar quantity since it is
expressed in terms of the work done.

STUDENT B: We were told in secondary school that the concept
of the potential of a field has no physical meaning. Only the
difference in the potentials of any two points of the field has a
physical meaning.

TEACHER: You are quite right. Strictly speaking, the preceding
discussion enables us to introduce precisely the difference in the
potentials; the potential difference between the two points a and
b of the field (denoted by (<p a — (py )) is defined as the ratio of the
work done by the forces of the field in moving charge q 0 from
point a to point b, to charge q 0 , i.e.

?a-?b=— (132)

%

Figure 89: Work done in an
electrostatic field.

However, if we assume that the field is absent at infinity (i.e.
cp oq = 0) equation (132) takes the form , then

%

(133)

HOW IS AN ELECTROSTATIC FIELD DESCRIBED?

227

In this manner, the potential of the field at the given point can be
determined in terms of the work done by the forces of the field
in moving a positive unit charge from the given point to infinity.
If the work is regarded as being done not by the field, but against
the forces of the field, then the potential at a given point is the
work that must be done in moving a positive unit charge from
infinity to the given point. Naturally, this definition rules out
experimental measurement of the potential at the given point
of the field, because we cannot recede to infinity in experiment.
Precisely for this reason it is said that the difference of the po¬
tentials of two points in the field has a physical meaning, while
the potential itself at some point has not. We can say that the
potential at a given point is determined with an accuracy to an
arbitrary constant. The value of the potential at infinity is com¬
monly taken as this constant. The potential is measured from
this value. It is assumed, for convenience, that the potential at
infinity equals zero.

Within the scope of these assumptions, the potential of a field,
set up by a point charge q x measured at a point a distance r, from
the charge, equals

?{r) = — ( 134 )

r

You should have no difficulty in determining the potential of a
field, set up by several point charges, at some point r.

STUDENT B: We shall denote the value of the potential at point
r due to each of the charges separately as <Pi(r), cp 2 (?), etc. The
total potential <p(r) is equal, evidently, to the algebraic sum of
the potentials from the separate charges. Thus

<p(r) = <Pi {?) + <p 2 (r) + ■ ■ ■ (I 35 )

In this summation, the potential from a positive charge is taken
with a plus sign and that from a negative charge with a minus
sign.

TEACHER: Quite correct. Now let us consider the concept of
equipotential surfaces. The locus of the points in a field having
the same potential is called an equipotential surface (or surface

228 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

of constant potential). One line of force and one equipotential
surface pass through each point in a field. How are they oriented
with respect to each other?

STUDENT B: I know that at each point the line of force and the
equipotential surface are mutually perpendicular.

TEACHER: Can you prove that?

STUDENT B: No, I probably can’t.

TEACHER: This proof is not difficult. Assume that the line
of force aa-y, and the equipotential surface S (Figure 90) pass
through a certain point a. The field intensity at point a is de¬
scribed by vector E a . Next we shall move charge q 0 from point
a to a certain point b which lies on the equipotential surface
S at a short distance A/ from point a. The work done in this
movement is expressed by the equation

A = F e Al cos a = E a q 0 Al cos a (136)

where a is the angle between vector E a and the direction of the
movement. This same amount of work can be expressed as the
difference in the field potentials at points a and b. Thus we can
write another relationship:

A = %(?<,-?b) (137)

Since both points a and b lie on the same equipotential surface,
then it follows that <p a = This means that according to equa¬
tion (137), the work A should be equal to zero. Substituting this
result into equation (136), we obtain

E a q 0 Alcosa = 0 (138)

Of all the factors in the left-hand side of equation (138), only
cos a can be equal to zero. Thus, we conclude that a = 90°. It
is clear to you, I think, that this result is obtained for various
directions of movement ab, provided these movements are
within the limits of the equipotential surface S. The curvature of
the surface does not impair our argument because the movement

Figure 90: Proving that lines of
force and equipotential surface
are mutually perpendicular.

HOW IS AN ELECTROSTATIC FIELD DESCRIBED?

229

A/ is very small. Along with the use of lines of force, cross-
sections of equipotential surfaces are employed to depict an
electrostatic field graphically. Taking advantage of the fact that
these lines and surfaces are mutually perpendicular, a family
of cross-sections of equipotential surfaces can be drawn from a
known family of lines of force, or vice versa.

(To STUDENT A) Will you try to draw the cross-sections of
equipotential surfaces for the case shown in (Figure 87(a))? To
avoid confusing them with the lines of force, draw the crosssec¬
tions of the surfaces with dashed lines.

STUDENT A: I shall draw the dashed lines so that they always
intersect the lines of force at right angles. Here is my drawing
(Figure 91).

Figure 91: Crosssections of
equipotential surfaces and lines
of force.

TEACHER: Your drawing is correct.

§ 24 How Do Lines Of Force Behave Near The
Surface Of A Conductor?

TEACHER: Let us introduce some conducting body into an elec¬
trostatic field. You know well that a conductor in a field is char¬
acterized by a quantity called the capacitance. But did you ever
ask yourself why we speak of the capacitance of a conductor, but
never of a dielectric?

STUDENT A: It never occurred to me.

TEACHER: How do you define the capacitance of an isolated
conductor?

STUDENT A: It is the quantity of electricity that must be im¬
parted to the conductor to increase its potential by one unit.

TEACHER: Mind you that you speak here of the potential as
being a characteristic of a body. But up till now the potential was
regarded as a characteristic of the field and, as such, it varied from
point to point. The potential is a function of the coordinates of
the corresponding point of a field. Can we speak of it as being a
characteristic of a body? If we can, then why?

STUDENT B: This is possible if the body is a conductor. The fact
is that all points of a conductor placed in an electrostatic field
have the same potential. A conductor is an equipotential body.

TEACHER: On what do you base your statement?

232 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: A conductor has free charges. Therefore, if a differ¬
ence in potential existed between any two points of the conduc¬
tor, there would be an electric current between these points. This
is obviously impossible.

TEACHER: Quite correct. It can be said that when a conductor
is brought into an electrostatic field, the free charges in the con¬
ductor are redistributed in such a manner that the field intensity
within the conductor becomes equal to zero. This actually sig¬
nifies that all the points of the conductor (both inside and on its
surface) have the same potential. The uniformity of the potential
at all points of a conductor enable us to speak of the potential
of the conductor as a body. I wish to point out that there are
no free charges in a dielectric and therefore no redistribution
of charges can occur. Incidentally, just how are the free charges
redistributed in a conductor?

STUDENT B: They are concentrated on its surface. The greater
the curvature of any projecting element of the conductor, the
denser the charges. The maximum charge density will be at a
sharp point.

TEACHER: Exactly. Now it is clear that a conductor in an elec¬
trostatic field is an equipotential body. It follows that the surface
of the conductor is an equipotential surface. On the basis of this
conclusion tell me how the lines of force of an electrostatic field
behave near the surface of a conductor?

STUDENT B: Since the lines of force are always perpendicular to
equipotential surfaces, they must “run” into the surface of the
conductor at right angles.

TEACHER: Unfortunately examinees frequently don’t know this.
You should have no difficulty in drawing a picture of the lines
of force in the field of a parallel-plate capacitor with a metal ball
between the plates. As a rule, examinees are greatly puzzled by
this question.

STUDENT B: The lines of force should approach the plates of the
capacitor and the surface of the ball at right angles. Thus, the

HOW DO LINES OF FORCE BEHAVE NEAR THE SURFACE OF A CONDUCTOR? 233

picture of the lines of force will resemble that shown in Figure 92.

Figure 92: Lines of force.

TEACHER: Everything is correct. I can’t understand why some
examinees think that the lines of force must bypass the ball.

Now let us consider the following problem. A point charge
+q is located at a distance r from the earth’s surface. It should
induce a charge of opposite sign in the earth. As a result, a force
of electric attraction is developed between the charge and the
earth. Find this force. I suggest that both of you think about this
problem.

STUDENT A: The charge induced in the earth should be equal to
the charge +q. It follows that the required force equals g 2 /r 2 .

STUDENT B: I don’t agree with this. STUDENT A assumed that
the charge induced in the earth is concentrated at one point
(point A in Figure 93 (a)). Actually, however, the induced charge
is not concentrated at one point but is distributed over the sur¬
face of the earth. For this reason, we know beforehand that the
required force must be less than g 2 /r 2 .

TEACHER: I fully agree with you. How then shall we go about
finding the force of attraction between the charge and the earth?

STUDENT B: It seems to me that we must examine the field be¬
tween the charge and the earth’s surface. The surface of the earth

234 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 93: Lines of force.

is evidently an equipotential one. Consequently, near the earth’s
surface the equipotential surfaces of the field must be close in
shape to planes. At the same time, the equipotential surfaces
in the vicinity of the charge must be spherical. This enables us
to draw a qualitative picture of the equipotential surfaces (or,
more exactly, of the cross-sections of these surfaces). When this
is done, we can draw the lines of force according to the familiar
rule. This has been done in Figure 93 (b), where the lines of force
are solid and the cross-sections of the surfaces are dashed lines.

TEACHER: Continue your line of reasoning, please. Doesn’t
your picture of the lines of force in Figure 93 (b) remind you of
something?

STUDENT B: Yes, of course. This picture certainly resembles the
one with the lines of force of two point charges that are equal
in magnitude and opposite in sign. I shall draw this picture at
the right (see Figure 93 (c)). Now everything is quite clear. In
both cases (see Figure 93 (b) and (c)), the appearance of the field
near charge +q is the same. According to equation (130) this
means the same force acts on charge +q in both cases. Thus, the
required force is q 2 /Ar 2 .

TEACHER: Your reasoning is faultless. This problem clearly

HOW DO LINES OF FORCE BEHAVE NEAR THE SURFACE OF A CONDUCTOR? 235

shows that the concept of the field may be exceptionally fruitful.

§ 25 How Do You Deal With Motion In A Uni¬
form Electrostatic Field?

TEACHER: Assume that a charged body moves in a uniform
electrostatic field, i.e. in a field where each point has the same
intensity E both in magnitude and direction. An example is the
field between the plates of a parallel-plate capacitor. Can you
see any resemblance between the problem on the motion of a
charged body in a uniform electrostatic field and any problems
considered previously?

STUDENT B: It seems to me that it closely resembles the problem
of the motion of a body in a gravitational field. Over relatively
short distances, the gravitational field of the earth can be re¬
garded as uniform.

TEACHER: Exactly. And what is the difference between motion
in an electrostatic field and in a gravitational field?

STUDENT B: Different forces act on the bodies. In an electro¬
static field, the force acting on the body is F e = E q (it imparts an
acceleration of a e = Eq/m to the body). The force in a gravita¬
tional field is P = mg (imparting the acceleration g to the body).
Here m is the mass of the body and q is its electric charge.

TEACHER: I wish that all examinees could understand the simple
truth that the motion of a body in any uniform field is kinemat¬
ically the same. What differs is only the value of the force acting
on the body in different fields. The motion of a charged body in

238 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

a uniform electrostatic field is of the same nature as the motion
of an ordinary stone in the earth’s field of gravitation. Let us
consider a problem in which the motion of a body takes place
simultaneously in two fields: gravitational and electrostatic. A
body of mass m with a charge +q is thrown upward at an an¬
gle of a, to the horizontal with an initial velocity v Q . The body
travels simultaneously in the field of gravitation and in a uniform
electrostatic field with an intensity E. The lines of force of both
fields are directed vertically downward (Figure 94 (a)). Find the
time of light 7) range Z.j and maximum height reached T/j.

(a)

Figure 94: Lines of force of two
fields in which a body travels.

STUDENT B: Two forces act on the body: the weight P = mg
and the electric force F e = Eq. In the given case, both forces are
parallel. As in § 5,1 can resolve the initial velocity vector ta) into
components in two directions ...

TEACHER: (interrupting): Just a minute! Do you want to repeat
the solution demonstrated in a similar problem in § 5?

STUDENT B: Yes, at least briefly.

HOW DO YOU DEAL WITH MOTION IN A UNIFORM ELECTROSTATIC FIELD?

239

TEACHER: There is no need to do that. You can refer directly to
the results in equations (15), (16) and (17). Just imagine that the
body travels in a “stronger” field of gravitation characterized by
a total acceleration equal to g + E(q/m). Make the following
substitution in equations (15), (16) and (17))

g + (— ) for g (139)

\m J

and you will obtain the required results at once:

7*1 =

2t? 0 sin a

V Eq

g + -

V m

Vq sin 2 a

g+ —

H,=

2 • 2
1 vt sin a

g+ i a )

(140)

(141)

(142)

STUDENT A: There is one point here that I don’t understand. In
comparison with the corresponding problem in § 5, an additional
force F e acts on the body in the given problem. This force is
directed vertically downward and therefore should not influence
the horizontal motion of the body. Why then, in the given case,
does it influence the range of flight Z.j?

TEACHER: The range depends upon the time of flight, and this
time is determined from a consideration of the vertical motion of
the body.

Now let us make a slight change in the conditions of the prob¬
lem: assume that the lines of force of the electrostatic field are
directed at an angle /3 to the vertical (Figure 94 (b)). As before,
find the time of flight T 2 , range L 2 and maximum height reached
H 2 .

STUDENT A: First I shall resolve force F e into two components:
vertical (F e cos j3) and horizontal (F e sin j3). This problem re-

240 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

minds me of the problem with the tail wind in § 5. Here the
component F e sin [3 plays the part of the “force of the wind”.

TEACHER: Quite right. Only remember that in contrast to
the problem with the tail wind you mentioned, here we have a
different vertical force, namely: mg + F e cos [3.

STUDENT A: I shall make use of equations (15), (16) and (18), in
which I’ll make the following substitutions

Eq cos (3 _

g + —£- !- for g

m

Eq sin B F

--- n f° r —

mg + Eq cos [3 P ,

(143)

After this I obtain the required results at once

t 2

h 2

2v 0 sin a

g +

/ Eqcos f3\

\ m J

? • 2

Vq sin a /

g +

/ Eqcosf3\ V

\ m )

1

2 • 2
sin a

g +

Eqcos [3

1 +

Eq sin [3 tan a
mg +£<7 cos fi

m

(144)

(145)

(146)

TEACHER: Absolutely correct. Unfortunately, examinees are
often incapable of drawing an analogy between motion in a
field of gravitation and motion in a uniform electrostatic field.
Consequently, such problems prove to be excessively difficult for
them.

STUDENT A: We did not study such problems before. The only
problem of this kind I have ever encountered concerns the mo¬
tion of an electron between the plates of a parallel-plate capac¬
itor, but we neglected the influence of the gravitational field
on the electron. I remember that such problems seemed to be
exceedingly difficult.

HOW DO YOU DEAL WITH MOTION IN A UNIFORM ELECTROSTATIC FIELD?

241

TEACHER: All these problems are special cases of the problem
illustrated in Figure 94 (a), since in the motion of an electron
inside a capacitor the influence of the gravitational field can be
neglected. Let us consider one such problem.

Having an initial velocity v 1 an electron flies into a parallel-
plate capacitor at an angle of a 1 and leaves the capacitor at an
angle of a 2 to the plates as shown in Figure 95. The length of the
capacitor plates is L. Find the intensity E of the capacitor field
and the kinetic energy of the electron as it leaves the capacitor.
The mass m and charge q of the electron are known.

Figure 95: Electron in a parallel
plate capacitor.

I denote by v 2 the velocity of the electron as it flies out of the
capacitor. Along the plates the electron flies at uniform velocity.
This enables us to determine the time of flight T inside the
capacitor

r =

L

v x COS ctj

The initial and final components of the electron velocity per¬
pendicular to the plates are related by the familiar kinematic
relationship for uniformly decelerated motion

Eq _

v 2 sin a 2 = v l sin a 1 - T

m

= v 1 sma 1 —

Eq

m Vl'iCosq : 1

from which, taking into account that the velocity component
along the plates remains unchanged ( v j cos a 1 = v 2 cos a 2 ), we

242 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

obtain

v 1 cos a 1 tan or, = v 1 sin a

Eq

L

m

v 1 cos a 1

From this equation we determine the intensity of the capacitor

field

mv\ cos 2 a 1

E = (tan a 1 — tan a 2 ) -

qL

(147)

The kinetic energy of the electron as it flies out of the capacitor
is

(148)

Is everything quite clear in this solution?

STUDENT A: Yes. Now I know how to solve such problems.

TEACHER: Also of interest are problems concerning the vibra¬
tion of a pendulum with a charged bob located within a parallel-
plate capacitor. We shall consider the following problem. A bob
of mass m with a charge q is suspended from a thin string of
length / inside a parallel-plate capacitor with its plates oriented
horizontally. The intensity of the capacitor field is E, and the
lines of force are directed downward (Figure 96 (a)). Find the
period of vibration of such a pendulum.

STUDENT B: Since in the given case the lines of force of the
electrostatic field and of the gravitational field are in the same
direction, I can use the result of equation (75) for an ordinary
pendulum after substituting the sum of the accelerations (g +
Eq/m ) for the acceleration of gravity g. Thus the required
period of vibration

T = 2n

l

(149)

TEACHER: Quite correct. As you see, the posed problem is very
simple if you are capable of using the analogy between motion in
a uniform electrostatic field and in a gravitational field.

HOW DO YOU DEAL WITH MOTION IN A UNIFORM ELECTROSTATIC FIELD?

243

Figure 96: Oscillations of a
charged bob in capacitor plates.

STUDENT A: Equation (149) resembles equation (77) in its struc¬
ture.

TEACHER: This is quite true. Only in equation (77) the addend to
the acceleration g was due to the acceleration of the frame of ref¬
erence (in which the vibration of the pendulum was investigated),
while in equation (149) the addend is associated with the presence
of a supplementary interaction.

How will equation (149) change if the sign of the charges on the
capacitor plates is reversed?

STUDENT A: In this case the period of vibration will be

(150)

TEACHER: Good. What will happen to the pendulum if we
gradually increase the intensity of the capacitor field?

STUDENT A: The period of vibration will increase, approaching
infinity at E = mg/q. If E continues to increase further, then we

244 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

will have to fasten the string to the lower and not the upper plate
of the capacitor.

TEACHER: What form will the equation for the period take in
this case?

STUDENT A: This equation will be of the form

/

T = 2n

(151)

TEACHER: Good. Now let us complicate the problem to some
extent. We will consider the vibration of a pendulum with a

charged bob inside a capacitor whose plates are oriented, not

horizontally, but vertically (Figure 96 (b)). In this case, the accel¬
erations g and ( Eq/m ) are directed at right angles to each other.
As before, find the period of vibration of the pendulum and, in

addition, the angle a that the string makes with the vertical when

the pendulum is in the equilibrium position.

STUDENT B: Taking into consideration the line of reasoning
given in the present section and in § 12,1 can conclude at once
that:

( 1 ) the period of vibration is expressed in terms of the effective
acceleration g ^ which is the vector sum of the accelerations
of the earth’s gravity and of the electrostatic field; and

( 2 ) the equilibrium direction of the string coincides with the
vector of the above-mentioned effective acceleration (this
direction is shown in Figure 96 (b) by a dashed line).

Thus

T = 2n

l

(152)

and

(153)

gm

HOW DO YOU DEAL WITH MOTION IN A UNIFORM ELECTROSTATIC FIELD?

245

TEACHER: Absolutely correct. I think that now it will be easy to
investigate the general case in which the capacitor plates make an
angle of [3 with the horizontal (Figure 96 (c)). The same problem
is posed: find the period of vibration of the pendulum and the
angle ex between the equilibrium direction of the pendulum
string and the vertical.

STUDENT B: As in the preceding case, the effective acceleration
is the vector sum of the acceleration of the earth’s gravity and
that of the electrostatic field. The direction of this effective
acceleration is the equilibrium direction of the pendulum string.
The effective acceleration g e ^ can be found by using the law of
cosines from trigonometry. Thus

Eq

Eq

g e ff = g — +2 g — cos/3

C JJ \ m / m

Then

/

V

g 2 +

(—)
\ m /

2 + 2g

f—)

\ m /

cos /3

The value of tan a can be found as follows

geffx Eq sin B

tan a = =----

geffy g m + Eq cos [3

(154)

(155)

TEACHER: Your answers are correct. Obviously, at [3 = 0, they
should lead to the results for the case of horizontal plates, and at
(3 = 90° to those for vertical plates. Please check whether this is
so.

STUDENT B: If f3 = 0, then cos (3 = 1 and sin /3 = 0. In this
case, equation (154) reduces to equation (149) and tana = 0 (the
equilibrium position of the string is vertical). If (3 = 90°, then
cos f3 = 0 and sin (3 = 1. In this case, equation (154) becomes
equation (152), and equation (155) reduces to equation (153).

TEACHER: I think that we have completely cleared up the prob¬
lem of the vibration of a pendulum with a charged bob inside a

246 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

parallel-plate capacitor. In conclusion, I want you to calculate
the period of vibration of a pendulum with a charged bob given
that at the point where the string of the pendulum is attached
there is another charge of exactly the same magnitude and sign
(Figure 97).

There are no capacitors whatsoever.

STUDENT A: According to Coulomb’s law, the bob will be re¬
pulsed from the point of suspension of the string with a force of
q 2 /l 2 . This force should impart an acceleration of q 2 /l 2 m to the
bob. The acceleration must be taken into account in the equa¬
tion for finding the period of vibration. As a result we obtain the
following expression

T

= In

(156)

Figure 97: A pendulum with
two charges: one at the bob
and other at the attachment
point.

TEACHER: (to STUDENT B): Do you agree with this result?

STUDENT B: No, I don’t. For equation (156) to be valid, it is
necessary for the acceleration g 2 / l 2 m to be directed vertically
downward at all times. Actually, it is so directed only when the
pendulum passes the equilibrium position. Thus it is clear that
equation (156) is wrong in any case. However, I don’t think that I
can give the right answer.

TEACHER: That you understand the error in equation (156)
is a good sign in itself. In the given case, the electric force is
at all times, directed along the string and is therefore always
counterbalanced by the reaction of the string. It follows that
the electric force does not lead to the development of a restoring
force and, consequently, cannot influence the period of vibration
of the pendulum.

STUDENT B: Does that mean that in the given case the period of
vibration of the pendulum will be found by equation (75) for a
pendulum with an uncharged bob?

HOW DO YOU DEAL WITH MOTION IN A UNIFORM ELECTROSTATIC FIELD? 247

TEACHER: Exactly. In the case we are considering, the field of
electric forces is in no way uniform and no analogy can be drawn
with a gravitational field.

Problems

45. An electron flies into a parallel-plate capacitor in a direction parallel
to the plates and at a distance of 4 cm from the positively charged
plate which is 15 cm long. How much time will elapse before the
electron falls on this plate if the intensity of the capacitor field
equals 500 V/m)? At what minimum velocity can the electron fly
into the capacitor so that it does not fall on the plate? The mass of
the electron is 9 x ICC 28 g, its charge is 4.8 x 10~ 10 esu (electrostatic
units).

46. An electron flies into a parallel-plate capacitor parallel to its plates
at a velocity of 3 x 10 6 ms. Find the intensity of the field in the
capacitor if the electron flies out of it at an angle of 30° to the plates.
The plates are 20 cm? long. The mass and charge of the electron are
known (see problem No. 45).

47. Inside a parallel-plate capacitor with a field intensity E, a bob with
a mass m and charge q, suspended from a string of length /, rotates
with uniform motion in a circle (Figure 98). The angle of inclina¬
tion of the string is a. Find the tension of the string and the kinetic
energy of the bob.

48. Two balls of masses m t and m 2 and with charges +qy and +q 2 are
connected by a string which passes over a fixed pulley. Calculate the
acceleration of the balls and the tension in the string if the whole
system is located in a uniform electrostatic field of intensity E
whose lines of force are directed vertically downward. Neglect any
interaction between the charged balls.

49. A ball of mass m with a charge of +q can rotate in a vertical plane
at the end of a string of length / in a uniform electrostatic field
whose lines of force are directed vertically upward. What horizontal
velocity must be imparted to the ball in its upper position so that

1 t

-T "

Figure 98: Pendulum inside a
parallel plate capacitor.

248 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

the tension of the string in the lower position of the ball is 10
the weight of the ball?

§ 26 Can You Apply Coulomb's Law?

TEACHER: Let us discuss Coulomb’s law in more detail, as well
as problems that are associated with the application of this law.
First of all, please state Coulomb’s law.

STUDENT A: The force of interaction between two charges is
proportional to the product of the charges and inversely propor¬
tional to the square of the distance between them.

TEACHER: Your statement of this law is incomplete; you have
left out some points.

STUDENT B: Perhaps I should add that the force of interaction
is inversely proportional to the dielectric constant K e of the
medium. Is that it?

TEACHER: It wouldn’t be bad to mention it, of course. But that is
not the main omission. You have forgotten again that a force is a
vector quantity. Consequently, in speaking of the magnitude of
a force, don’t forget to mention its direction (in this connection,
remember our discussion of Newton’s second law in § 4).

STUDENT A: Now I understand. You mean we must add that the
force with which the charges interact is directed along the line
connecting the charges?

TEACHER: That is insufficient. There are two directions along a
line.

250 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT A: Then we must say that the charges repulse each
other if they are of the same sign and attract each other if they
are of opposite signs.

TEACHER: Good. Now, if you collect all these additions, you will
obtain a complete statement of Coulomb’s law. It would do no
harm to emphasize that this law refers to interaction between
point charges.

STUDENT B: Can the equation of Coulomb’s law be written
so that it contains full information concerning the law? The
ordinary form

F

b T7*

e

(157)

yields no information on the direction of the force.

TEACHER: Coulomb’s law can be written in this way. For this we
first have to find out what force we are referring to. Assume that
we mean the force with which charge q l acts on charge q 2 (and
not the other way round). We introduce coordinate axes with the
origin at charge q l . Then we draw vector from the origin to the
point where charge q 1 is located (Figure 99). This vector is called
the radius vector of charge q 2 . In this case, the complete formula
of Coulomb’s law will be

B=B ir¥ ( 158 )

K p r*

where factor B depends upon the selection of the system of units.

STUDENT A: But in this equation the force is inversely propor¬
tional, not to the square, but to the cube of the distance between
the charges!

TEACHER: Not at all. Vector (r/r) is numerically equal to unity
(dimensionless unity!). It is called a unit vector. It serves only to
indicate direction.

STUDENT A: Do you mean that I can just write equation (158) if
I am asked about Coulomb’s law? Nothing else?

Z

Figure 99: Direction in
Coulomb’s law.

CAN YOU APPLY COULOMB’S LAW? 251

TEACHER: You will only have to explain the notation in the
equation.

STUDENT A: And what if I write equation (157) instead of (158)?

TEACHER: Then you will have to indicate verbally the direction
of the Coulomb force.

STUDENT A: How does equation (158) show that the charges
attract or repulse each other?

TEACHER: If the charges are of the same sign, then the product
q^q 2 is positive. In this case vector F is parallel to vector r.
Vector F is the force applied to charge q 2 ', charge q~> is repulsed
by charge q 1 . If the charges are of opposite sign, the product q l q 1
is negative and then vector F will be anti-parallel to vector r i.e.
charge q 1 will be attracted by charge q l .

STUDENT A: Please explain what we should know about factor

B.

TEACHER: This factor depends upon the choice of a system
of units. If you use the absolute electrostatic (cgse) system

of units, then B = 1; if you use the International System

1

of Units (51), then B = -, where the constant f 0 =

47re 0

8.85 x lCT 12 C 2 /(Nm 2 )

Let us solve a few problems on Coulomb’s law.

Problem 1 . Four identical charges q are located at the corners of a
square. What charge Q of opposite sign must be placed
at the centre of the square so that the whole system of
charges is in equilibrium?

STUDENT A: Of the system of five charges, four are known and
one is unknown. Since the system is in equilibrium, the sum
of the forces applied to each of the five charges equals zero. In
other words, we must deal with the equilibrium of each of the
five charges.

Figure 100: What is the charge
of Q?

-*b8

252 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

TEACHER: That will be superfluous. You can readily see that
charge Q is in equilibrium, regardless of its magnitude, due to its
geometric position. Therefore, the condition of equilibrium with
respect to this charge contributes nothing to the solution. Owing
to the symmetry of the square, the remaining four charges q are
completely equivalent. Consequently, it is sufficient to consider
the equilibrium of only one of these charges, no matter which.
We can select, for example, the charge at point A (Figure 100).
What forces act on this charge?

STUDENT A: Force iq from the charge at point B, force F 2 from
the charge at point D and, finally, the force from the sought-for
charge at the centre of the square.

TEACHER: I beg your pardon, but why-didn’t you take the charge
at point C into account?

STUDENT A: But it is obstructed by the charge at the centre of
the square.

TEACHER: This is a naive error. Remember: in a system of
charges each charge is subject to forces exerted by all the other
charges of the system without exception. Thus, you will have to
add force T 3 acting on the charge at point A from the charge at
point C. The final diagram of forces is shown in Figure 100.

STUDENT A: Now, everything is clear. I choose the direction AS
and project all the forces applied to the charge at point A on this
direction. The algebraic sum of all the force projections should
equal zero, i.e.

F 4 = 2 F 1 cos 45° + T 3

Denoting the side of the square by a, we can rewrite this equa¬
tion in the form

from which

Q = | (2V2 + l)

(159)

CAN YOU APPLY COULOMB’S LAW?

253

TEACHER: Quite correct. Will the equilibrium of this system of
charges be stable?

STUDENT B: No, it won’t. This is unstable equilibrium. Should
anyone of the charges shift slightly, all the charges will begin
moving and the system will break up.

TEACHER: You are right. It is quite impossible to devise a stable
equilibrium configuration of stationary charges.

Problem 2. Two spherical bobs of the same mass and radius,
having equal charges and suspended from strings
of the same length attached to the same point, are
submerged in a liquid dielectric of permittivity K e ,
and density p 0 . What should the density p of the
bob material be for the angle of divergence of the
strings to be the same in the air and in the dielectric?

Figure 101: What is the charge
of Q?

STUDENT B: The divergence of the strings is due to Coulomb
repulsion of the bobs. Let F el denote Coulomb repulsion in the
air and F el in the liquid dielectric.

TEACHER: In what way do these forces differ?

254 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: Since, according to the conditions of the problem,
the angle of divergence of the strings is the same in both cases,
the distances between the bobs are also the same. Therefore, the
difference in the forces F gl and F e2 is due only to the dielectric
permittivity. Thus

F e \ = K e F e2 ( 160 )

Let us consider the case where the bobs are in the air. From the
equilibrium of the bobs we conclude that the vector sum of the
forces F el and the weight should be directed along the string
because otherwise it cannot be counterbalanced by the reaction
of the string (Figure 101 (a)). It follows that

Pel

P

= tan a

where a is the angle between the string and the vertical. When
the bobs are submerged in the dielectric, force F el should be
replaced by force and the weight P by the difference (P —
Fy), where Fy is the buoyant force. However, the ratio of these
new forces should, as before, be equal to tana (Figure 101 (b)).
Thus

- el

P-Fu

= tana

Using the last two equations, we obtain

■ el

1 el

P-F u

After substituting equation (160) and taking into consideration
that P = Vgp and Fy = Vgp 0 , we obtain

K e 1
P P Po

and the required density of the bob material is

Po K e

p =

K e ~ 1

(161)

TEACHER: Your answer is correct.

CAN YOU APPLY COULOMB’S LAW?

255

Problem 3. Two identically charged spherical bobs of mass m
are suspended on strings of length / each and at¬
tached to the same point. At the point of suspension
there is a third ball of the same charge (Figure 102).
Calculate the charge q of the bobs and ball if the
angle between the strings in the equilibrium position
is equal to a.

STUDENT B: We shall consider bob A. Four forces (Figure 102)
are applied to it. Since the bob is in equilibrium, I shall resolve
these forces into components in two directions ...

TEACHER: (interrupting): In the given case, there is a simpler
solution. The force due to the charge at the point of suspension
has no influence whatsoever on the equilibrium position of the
string: force Pel acts along the string and is counterbalanced in
any position by the reaction of the string. Therefore, the given
problem can be dealt with as if there were no charge at all at the
point of suspension of the string. As a rule, examinees don’t
understand this.

STUDENT B: Then we shall disregard force Pel- Since the vector
sum of the forces F el and P must be directed along the string, we
obtain

TEACHER: Note that this result does not depend upon the pres¬
ence or absence of a charge at the point of suspension.

STUDENT B: Since

*el =

4/ 2 sin 2

we obtain from equation (162):

< 7 2

4 l 2 mg sin 2 ^

a

= tan —
2

Figure 102: What is the charge
of Q?

256 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Solving for the required charge, we obtain

<7 = 2/sin (f) ^/wgtan| ( 163 )

TEACHER: Your answer is correct.

STUDENT A: When will the presence of a charge at the point of
suspension be of significance?

TEACHER: For instance, when it is required to find the tension of
the string.

Problems

50. Identical charges are +q located at the vertices of a regular hexagon.
What charge must be placed at the centre of the hexagon to set the
whole system of charges at equilibrium?

51. A spherical bob of mass m and charge q suspended from a string
of length / rotates about a fixed charge identical to that of the bob
(Figure 103). The angle between the string and the vertical is a.
Find the angular velocity of uniform rotation of the bob and the
tension of the string.

52. A spherical bob of mass m with the charge q can rotate in a vertical
plane at the end of a string of length /. At the centre of rotation
there is a second ball with a charge identical in sign and magnitude
to that of the rotating bob. What minimum horizontal velocity
must be imparted to the bob in its lower position to enable it to
make a full revolution?

Figure 103: What is the angular
velocity of uniform rotation of
the bob and the tension of the
string.

Electric currents have become an integral part of our everyday life, and
so there is no need to point out the importance of the Ohm and the
Joule-Lenz laws. But how well do you know these laws?

§27 Do You Know Ohm's Law?

TEACHER: Do you know Ohm’s law?

STUDENT A: Yes, of course. I think everybody knows Ohm’s
law. This is probably the simplest question in the entire physics
course.

TEACHER: We shall see. A Portion of electric circuit is shown
in Figure 104 (a). Here $ is the electromotive force ( emf ) and it
is directed to the right; R { and R 2 are resistors; r is the internal
resistance of the seat of the electromotive force; and (p A and cp B
are the potentials at the ends of the given portion of the circuit.
The current flows from left to right. Find the value I of this
current.

(a)

A

?A° -C

(b) A

%° -

(CJ M

l r

+

?A C

s

+ !■-

'w

Rz

B

B

B

Figure 104: Find the value of
the current in the circuit.

STUDENT A: But you have an open circuit!

TEACHER: I proposed that you consider a portion of some large
circuit. You know nothing about the rest of the circuit. Nor do

260 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

you need to, since the potentials at the end of this portion are
given.

STUDENT A: Previously we only dealt with closed electric cir¬
cuits. For them the Ohm’s law can be written in the form:

I =

£

R + r

(164)

TEACHER: You are mistaken. You also considered elements of
the circuit. According to Ohm’s law, the current in an element of
a circuit is equal to the ratio of the voltage and the resistance.

STUDENT A: But is this a circuit element?

TEACHER: Certainly. One such element is illustrated in Fig¬
ure 104 (b). For this element you can write Ohm’s in the form

I= ?A~?B

R

(165)

Instead of potential difference {cp A — cp B ) between the ends of the
element, you previously employed the simpler term “voltage”,
denoting it by the letter V.

STUDENT A: In any case, we did not deal with an element of
circuit of the form shown in Figure 104 (a).

TEACHER: Thus, we find that you know Ohm’s law for the spe¬
cial cases of a closed circuit and for the simplest kind of element
which includes no emf. You do not, however, know Ohm’s law
for the general case. Let us look into this together.

Figure 105 (a) shows the change in potential along a given por¬
tion in the circuit. The current flows form left to right and
therefore the potential drops from A to C. The drop in poten¬
tial across the resistor R { is equal to IR l . Further, we assume
that the plates of a galvanic cell are located at C and D. At these
points upwards potential jumps occur, the sum of the jumps is
the emf equal to $. Between C and D the potential drops across
the internal resistance of the cell; the drop on potential across
the resistor R 2 equals IR 2 . The sum of the drops across all the

DO YOU KNOW OHM’S LAW? 261

resistances of the portion minus the upward potential jump is
equal to V. It is the potential difference between the ends of the
portion being considered. Thus

I{R l + R 2 + r ) - S' = (p A ~ <Pb

m (b)

Figure 105: Find the value of
the current in the circuit.

From this we obtain the expression for the current, i.e. Ohm’s
law for the given portion of the circuit

& + ( C PA ~ 9b)
Ri + R-2 ^

(166)

Note that from this last equation we can readily obtain the spe¬
cial cases familiar to you. For the simplest element containing no
emf we substitute § = 0 and r = 0 into equation (166)). Then

I= ?A-<Pb
ft j ■

which corresponds to equation (165). To obtain a closed circuit,
we must connect the ends A and B of our portion. This means
that (j>A = </>b- Then

(i?i +R 2 + r)

262 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

This corresponds to equation (164).

STUDENT A: I see now that I really didn’t know Ohm’s law.

TEACHER: To be more exact, you knew it for special cases only.
Assume that a voltmeter is connected to the terminals of the cell
in the portion of circuit shown in Figure 104 (a). Assume also
that the voltmeter has sufficiently high resistance so that we can
disregard the distortions due to its introduction into the circuit.
What will the voltmeter indicate?

STUDENT A: I know that a voltmeter connected to the terminals
of a cell should indicate the voltage drop across the external
circuit. In the given case, however we know nothing about the
external circuit.

TEACHER: A knowledge of external circuit is not necessary for
our purpose. If the voltmeter is connected to points C and D,
it will indicate the difference in potential between these points.
You understand this, don’t you?

STUDENT A: Yes, of course.

TEACHER: Now look at Figure 105 (a). It is evident that the
difference in potential between points C and D equals ($ = Ir).
Denoting the voltmeter reading by V, we obtain the formula

V = S-Ir (167)

I would advise you to use this very formula since it requires no
knowledge of any external resistances. This is specially valuable
in cases when you deal with a more or less complicated circuit.
Note that equation (167) lies at the basis of a well known rule: If
the circuit is broken and no current flows 1 = 0, then V = S.
Here the voltmeter reading coincides with the value of the emf.

Do you understand all this?

STUDENT A: Yes, now it is clear to me?

TEACHER: As a check I shall ask you a question which exami¬
nees quite frequently find it difficult to answer. A closed circuit

DO YOU KNOW OHM’S LAW?

263

consists of n cells connected in series. Each element has an emf
£ and internal resistance r . The resistance of the connecting
wires is assumed to be zero. What will be the reading in the volt¬
meter connected to the terminals of one of the cells. As usual it is
assumed that no current passes through the voltmeter.

STUDENT A: I shall reason as in the preceding explanation. The
voltmeter reading will be V = S — I r . From Ohm’s law for the
given circuit we can find the current

n£ £

nr r

Substituting this in the first equation we obtain
v (T) r-a

Thus, in this case the voltmeter will read zero.

TEACHER: Absolutely correct. Only please remember that this
case was idealized. On one hand, we neglected the resistance of
the connecting wires, and on the other we assumed the resistance
of the voltmeter to be infinitely large, so don’t try to check this
result by experiment.

Now let us consider a case when the current in a portion of a
circuit flows in one direction and the emf acts in the opposite
direction. This is illustrated in Figure 104 (c). Draw a diagram
showing the change of potential along this portion.

STUDENT A: Is it possible for the current to flow against the emf?

TEACHER: You forget that we have here only the portion of a
circuit. The circuit may contain other emf’s outside the portion
being considered, under whose effect the current in this portion
may flow against the given emf.

STUDENT A: I see. Since the current flows from the left to right,
there is a potential drop equal to IR X from A to C. Since the
emf is now in the opposite direction, the potential jumps at
the points C and D should now reduce the potential instead of

264 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

increasing it. From point C to point D the potential should drop
by the amount I r, and from point D to point B, by IR 2 ■ As a
result we obtain the diagram of Figure 105 (b).

TEACHER: And what form will Ohm’s law take in this case?

STUDENT A: It will be of the form

/= {<Pa-?b)~ s

R^ + R 2 H - t

(168)

TEACHER: Correct. And what will voltmeter indicate now?
STUDENT A: It can be seen from Figure 105 (b) that in this case

V = d? + Ir (169)

TEACHER: Exactly. Now consider the following problem. In the
electrical circuit illustrated in Figure 106, r = IQ, R = 100
and the resistance of the voltmeter R v = 200 Cl. Compute the
relative error of the voltmeter reading obtained assuming that the
voltmeter has infinitely high resistance and consequently causes
no distortion in the circuit.

We shall denote the reading of the real voltmeter by V and that
of the voltmeter with infinite resistance by V x . Then the relative
error would be

/

VC-V

VC

= 1-

V

(170)

Further, we shall take into consideration that

Ko =

S

R + r

R

(171)

and

V =

g

( RRy \

\R + R V )

( RRv )

\R+R V J

—©

Figure 106: Find the value of
the current in the circuit.

(172)

DO YOU KNOW OHM’S LAW?

265

After substituting equations (171) and 172 into (170) we obtain:

f Ry(R + r )

( R+R v )r+RR v

x Ry(R + r)

(r +R)R V + rR

= 1 ~l + f rR )

\(r +R)R V J

Since R v » R and R > r, the fraction in the denominator of the
last equation is much less than unity. Therefore, we can make use
of an approximation formula which is always useful to bear in
mind

(1 + d)" ss 1 + ceX (173)

This formula holds true at X « 1 for any value of a (whole
or fractional, positive or negative). Employing approximation
formula (173) with

a = — 1 and X

rR

0 r+R)Ry

we obtain

f =

rR

{r+R)Ry

(174)

Substituting the given numerical values into equation (174), we

1

find that the error is f & -= 0.0045.

y 220

STUDENT A: Does this mean that higher resistance of the volt¬
meter in comparison with the external resistance, the lower the
relative error, and that the more reason we have to neglect the
distortion of the circuit when the voltmeter is connected into it?

TEACHER: Yes, that’s so. Only keep in mind that Ry » A is a
sufficient, but not necessary condition for the smallness of the
error/. It is evident from equation (174) that error/ is small
when the condition Ry » r is complied with, i.e. the resistance
of the voltmeter is much higher than the internal resistance of
the current source. The external resistance in this case maybe be
infinitely high.

266 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Try to solve the following problem: In the electrical circuit
shown in Figure 107 (a), E = 6 V, r = 2/3 0, R = 20. Compute
the voltmeter reading.

(a)

b 2

Figure 107: Find the value of
the voltmeter reading.

STUDENT A: Can we assume that the resistance of the voltmeter
is infinitely high?

TEACHER: Yes, and the more so because this resistance is not
specified in the problem.

STUDENT A: But then, will the current flow through the resistors
in the middle of the circuit? It will probably flow directly along
the elements A t A 2 and SjS 2 -

TEACHER: You are mistaken. Before dealing with the currents,

I would advise you simplify the diagram somewhat. Since the
elements A X A 2 andffjS 2 have no resistance, it follows that <p A1 =
(p A2 and cp B j = tp B2 . Next, we can make use of the rule: if in a
circuit any two points have the same potential, without changing
the currents through the resistors. Let us apply this rule to our
case by making point A\ coincide with point A 2 , point with

DO YOU KNOW OHM’S LAW?

267

B 2 - We then obtain the diagram shown in Figure 107 (b). This
one is quite easy to handle. Therefore, I will give you the final
answer directly: the voltmeter reading will be 4 V. I shall leave
the necessary calculations to you as a home assignment.

Problems

53. An ammeter, connected into a branch of the circuit shown in Figure
108, has a reading of 0.5 A. Find the current through resistor R 4
if the resistances are: R j = 20, R 2 = 40, R 2 = lO, R 4 =

2 O and i? 5 = 1 O.

Figure 108: Find the value of
the current through resistor

r a .

54. In the electric circuit shown in Figure 109, S = 4 V, r = 1 O and R =
2 O. Find the reading of the ammeter.

Figure 109: Find the reading of
the ammeter.

55. The resistance of a galvanometer equals 0.2 O. Connected in parallel
to the galvanometer is a shunt with a resistance of 0.05 O. What
resistance should be connected in series with this combination to
make the total resistance equal to that of the galvanometer?

268 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

56. A voltmeter with a resistance of 100 Q is connected to the terminals
of a cell with an emf of 10 V and internal resistance of 1 fi. Deter¬
mine the reading of the voltmeter and compute the relative error of
its reading assuming that its resistance is infinitely high.

57. An ammeter with a resistance of 1 Q is connected into a circuit
with an external resistance of 49 Q and with a current source having
an emf of 10 V and an internal resistance of 1 Q. Determine the
reading of the ammeter and compute the relative error of its reading
assuming that it has no resistance.

§ 28 Can A Capacitor Be Connected Into A Direct-
Current Circuit ?

TEACHER: Let us consider the following problem. In the circuit
shown in Figure 110 C is the capacitance of the capacitor. Find
the charge Q on the capacitor plates if the emf of the current
source is S and its internal resistance is r.

STUDENT A: But can we use capacitor in a direct current circuit?
Anyway, no current will flow through it.

TEACHER: What if it doesn’t? But it will flow in the parallel
branches.

STUDENT A: I think I understand now. Since the current doesn’t
flow through the capacitor in the circuit of (Figure 110) , it will
not flow through resistor R 1 either. In the external part of the
circuit the current will only flow through resistor R 2 . We can
find the current through it by using the relation I = £/{R 2 + r)
and then the potential difference between points A and B will
equal the drop in voltage across the resistor R~> i.e.

?A~fB= IR 2 = 4~Z~ (175)

I don’t know what to do next. To find the charge on the capaci¬
tor plates, I must find the potential difference between points A
and F.

TEACHER: You were correct in concluding that no current flows
through resistor R 1 . In such a case, however all points of the

Figure 110: Find the reading of
the ammeter.

270 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

resistor should have the same potential (remember discussion
in § 24). That means that (p A = cp B . From this, making use of
equation (175); we find the required charge

C(aR 1

R 2 + r

(176)

Now consider the following problem. In the electric circuit
shown in Figure 111, £ = 4V, r = lfl, Rj = 3 C1,R 2 =

2 0, Cj = 2 ]iF, C 2 = 8 ]iF, C 3 = 4 pF, C 4 = 6]iF. Find the
charge on the plates of the capacitor.

C ! C 3

rlh

_ Rp

n * f

Hh

' c* 1

£-1 b--

' C 4

HH

R,

HH

Figure 111: Find the charge on
the plates of the capacitor.

In this connection recall the rules of adding capacitors connected
in series and parallel.

STUDENT A: I remember those rules. When capacitors are con¬
nected in parallel, their combined capacitance is simply the sum
of individual capacitances, i.e.

C = C 1 + C 2 + C 3 + C 4 + ... (177)

and when they are connected in series, the combined capaci¬
tance is given by the reciprocal of the sum of the reciprocals of
individual capacitances. Thus

11 1 1

c “ q q q'"

(178)

TEACHER: Exactly. Now making use of rule (177), we find the
capacitance between points A and B, and points F and D as well:

C AB = 2pF + 8pF = lOpF
C FD = 4pF + 6pF = lOpF

CAN A CAPACITOR BE CONNECTED INTO A DIRECT-CURRENT CIRCUIT? 271

The difference in the potential between points A and D is equal
to the voltage drop across resistor R t . Thus

?d-?a = ir l

R x + r

= 3 V

Obviously, resistor R 2 plays no part in the circuit and can be
ignored. Since C AB = C FD , then

3 V

( Pb- ( Pa = ( Pd~ c Pf = -^ = L5 v

Finally we can obtain the required charges:

Qi = C 1 (cp B -(p A ) = 3iiC

Q 2 = C 2 ((Pb-<Pa) = u V C
Q3 = C 3 (<p D - <p F ) = 6]iC
Q 4 = C A {cp D — <Pf) = 9 fC

Problems

58. In the circuit Figure 112, £ = 5V, r = 1 O, R 2 = 40,7?) =

3 O, and C = 3 pF. Find the charge on the plates of each capacitor.

R,

Figure 112: Find the charge on
the plates of each capacitor.

59. All the quantities indicated on the diagram of the circuit shown
in Figure 113 being known, find the charge on the plates of each
capacitor.

272 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

60. A parallel-plate capacitor with plates of length / is included in a
circuit as shown in Figure 114. Given are the emf S of the current
source, its internal resistance r and the distance d between the
plates. An electron with a velocity v 0 flies into the capacitor, paral¬
lel to the plates. What resistance R should be connected in parallel
with the capacitor so that the electron flies out of the capacitor at
an angle a to the plates? Assume the mass m and the charge q of the
electron to be known.

61. Two identical and mutually perpendicular parallel-plate capaci¬
tors, with plates of length / and a distance d between the plates are
included in the circuit shown in Figure 115. The emf S and the
resistance r of the current source are known. Find the resistance
R at which an electron flying at a velocity of v 0 into one of the ca¬
pacitors, parallel to its plates, flies into the second capacitor and
then flies out parallel to its plates. The mass m and charge q of the
electron are known.

62. A parallel-plate capacitor with plates of length / and a distance d
between them is included in the circuit shown in Figure 116 (the
emf S and the resistances R and r are known). An electron flies
into the capacitor at a velocity v 0 parallel to the plates. At what

Figure 113: Find the charge on
the plates of each capacitor.

Figure 114: Find the resistance
R.

Figure 115: Find the resistance
R.

CAN A CAPACITOR BE CONNECTED INTO A DIRECT-CURRENT CIRCUIT? 273

angle to the plates will the electron fly out of the capacitor if m and
q are known.

5 +

i

_1

+2£

■T

R ZR

1

iU

|

zr

■IrVl

Figure 116: At what angle to
the plates will the electron fly
out of the capacitor.

§ 29 Can you compute the resistance of a branched
portion of a circuit?

TEACHER: Compute the resistance of the portion of the circuit
shown in Figure 117 (a). You can neglect the resistance of wires
(leads).

Figure 117: At what angle to
the plates will the electron fly
out of the capacitor.

STUDENT A: If the resistance of the wires can be neglected, then
the leads can be completely disregarded. The required resistance

276 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

equals 3 R.

TEACHER: You answered without thinking. To neglect the re¬
sistance of the wire and to neglect the leads are two entirely
different things (though many examinees suppose them to be the
same). To throw a lead out of the circuit means to replace it with
an infinitely high resistance. Here on the contrary the resistance
of the leads equals zero.

STUDENT A: Yes, of course, I simply didn’t give it any thought.
But now I shall reason in the following manner. At point A the
current will be divided into two currents whose directions I have
shown in Figure 117 (b) by arrows. Here the middle resistor can
be completely disregarded and the total resistance is R/2.

TEACHER: Wrong again! I advise you to use the following rule:
find the points in the circuit with the same potential and then
change the diagram so that those points coincide with one an¬
other. The currents in the various branches of the circuit will
remain unchanged, but the diagram maybe substantially sim¬
plified. I have already spoken about this in § 27. Since in the
given problem the resistances of the leads equal zero, points A
and Aj have the same potential. Similarly points B and B 1 have
the same potential. In accordance with the rule I mentioned, we
shall change the diagram so that points with the same potential
will finally coincide with one another. For this purpose, we shall
gradually shorten the lengths of the leads. The consecutive stages
of this operation are illustrated in Figure 117 (c). As a result we
find that the given connection corresponds to an arrangement
with three resistors connected in parallel. Hence, the total resis¬
tance of the portion is R/ 3.

STUDENT A: Yes, indeed. It is quite evident from Figure 117 (c)
that the resistors are connected in parallel.

TEACHER: Let us consider the following example. We have a
cube made up of leads, each having resistance R Figure 118 (a).
The cube is connected into a circuit as shown in the diagram.
Compute the total resistance of the cube.

CAN YOU COMPUTE THE RESISTANCE OF A BRANCHED PORTION OF A CIRCUIT? 277

We can start by applying the rule I mentioned above. Indicate
the points having the same potential.

STUDENT A: I think that the three points A, A l and7l 9 will have
the same potential (see Figure 118 (a)) since the three edges of the
cube ( DA,DA { and DA 2 ) are equivalent in all respects.

(a)

TEACHER: Yes, and so are edges BC ,SCj and BC 2 ■ Therefore,
points C, and C 2 will have the same potential. Next, let us tear
apart our wire cube at the indicated points and, after bending the
edge wires, connect them together’ again so that points with the
same potential coincide with one another. What will the diagram
look like now?

STUDENT A: We shall obtain the diagram shown in Figure 118 (b).

TEACHER: Exactly. The diagram obtained in Figure 118 (b) is
equivalent to the initial diagram (with the cube) but is apprecia¬
bly simpler. Now you should have no difficulty in computing
the required total resistance.

1 1 1 5

-R+ -R + -R = -R

3 6 3 6

Figure 118: What is the total
resistance of the cube?

STUDENT A: It equals

278 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: How would you find the total resistance of a wire
figure in the form of a square with diagonals, connected into a
circuit as shown in Figure 119 (a)?

Figure 119: What is the total
resistance of the cube?

TEACHER: Again we must search for points with the same poten¬
tial. In the given case we readily see that the diagram has an axis
of symmetry which I shall indicate in Figure 119 (a) as a dashed
line. It is clear that all points lying on the axis of symmetry
should have the same potential which is equal to one half the sum
of the potentials of points A and D. Thus the potentials of points
O, Oj and 0 9 are equal to one another. According to the rule,
we can make these three points coincide with one another. As a
result, the combination of resistances is broken down into two
identical portions connected in series. One of these is shown in
Figure 119 (b). It is not difficult to compute the resistance of this
portion. If each of the wires, or leads, in the square has the same
resistance 7?, then the total resistance of the portion is (4/5)7?.
Thus the required total resistance of the square equals (8/5)7?.

STUDENT A: Do you mean to say that the main rule is to find
points on the diagram with the same potential and to simplify

CAN YOU COMPUTE THE RESISTANCE OF A BRANCHED PORTION OF A CIRCUIT? 279

the diagram by making these points coincide?

TEACHER: Exactly. In conclusion, I wish to propose an example
with an infinite portion. We are given a circuit made up of an
infinite number of repeated sections with the resistors R j and R 2
(Figure 120 (a)). Find the total resistance between points A and B.

Figure 120: What is the
total resistance of the given
configuration?

STUDENT A: Maybe we should make use of the method of math¬
ematical induction? First we will consider one section, then two
sections, then three, and so on. Finally we shall try to extend the
result to n sections for the case when n —» go.

TEACHER: No, we don’t need the method of mathematical in¬
duction here. We can begin with the fact that infinity will not
change if we remove one element from it. We shall cut the first
section away from the diagram (along R the dashed line shown in
Figure 120 (a)). Evidently, an infinite number of sections will still
remain and so the resistance between points C and D should be
equal to the required total resistance R. Thus the initial diagram
can be changed to the one shown in Figure 120 (b). The portion
of the circuit shown in Figure 120 (b) has a total resistance of

RR 2

R, + -—

1 (* + * 2 )

280 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Since this portion is equivalent to the initial portion of the cir¬
cuit, its resistance should equal the required resistance R. Thus
we obtain

R = R X +

RR 2

(R+R 2 )

i.e. a quadratic equation with respect to R:

R 2 -RR l -R 1 R 2 =0

Solving this equation we obtain

(179)

STUDENT A: Well, that certainly is an interesting method of
solving the problem.

Problems

63. In the electrical circuit shown in Figure 121, S = 4 V, r = 1Q and R =
45 Q. Determine the readings of the voltmeter and ammeter.

Figure 121: What are the
readings of the voltmeter and
ammeter?

64. Find the total resistance of the square shown in Figure 119 (a) as¬
suming that it is connected into the circuit at points A and C.

65. A regular hexagon with diagonals is made of wire. The resistance of
each lead is equal to R. The hexagon is connected into the circuit as
shown in Figure 122 (a). Find the total resistance of the hexagon.

66. Find the total resistance of the hexagon of Problem 65 assuming
that it is connected into the circuit as shown in Figure 122 (b).

CAN YOU COMPUTE THE RESISTANCE OF A BRANCHED PORTION OF A CIRCUIT? 281

Compute the total resistance of the hexagon given in Problem
65 assuming that it is connected into the circuit as shown in Fig¬
ure 122 (c).

Figure 122: What is the total
resistance?

§30 Why Did The Electric Bulb Burn Out?

STUDENT A: Why does an electric bulb burn out? From exces¬
sive voltage or from excessive current?

TEACHER: How would you answer this question?

STUDENT A: I think it is due to the high current.

TEACHER: I don’t much like your answer. Let me note, first, that
the question, as you put it, should be classified in the category of
provocative or tricky questions. An electric bulb burns out as a
result of the evolution of an excessively large amount of heat in
unit time, i. e. from a sharp increase in the heating effect of the
current. This, in turn, may be due to a change in any of various
factors: the voltage applied to the bulb, the current through the
bulb and the resistance of the bulb. In this connection, let us
recall all the formulas you know for finding the power developed
or expended when an electric current passes through a certain
resistance R.

STUDENT B: I know the following formulas:

P = (cp 1 - cp 2 )I

(180)

P = I 2 R

(181)

p _ ifi _ 9i)~

R

(182)

where P is the power developed in the resistance R, ( cp j — cp 2 )
is the potential difference across the resistance R and / is the
current flowing through the resistance R.

284 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT A: We usually used only formula (181), which ex¬
presses the power in terms of the square of the current and the
resistance.

TEACHER: It is quite evident that the three formulas are equiv¬
alent since one can be transformed into the others by applying
Ohm’s law. It is precisely the equivalence of the formulas that
indicates that in solving our problem we should not deal with the
current or voltage separately. We should take all three quantities
- the current, voltage and resistance - into account together. (To
STUDENT A): By the way, why do you prefer formula (181)?

STUDENT A: As a rule, the voltage supplied to a bulb is constant.
Therefore, the dependence of the power on the voltage is of no
particular interest. Formula (181) is the most useful of the three.

TEACHER: You are wrong in assigning a privileged position to
formula (181). Consider the following problem. The burner
of an electric table stove is made up of three sections of equal
resistance. If the three sections are connected in parallel, water
in a tea-kettle begins to boil in 6 minutes. When will the same
mass of water in the tea-kettle begin to boil if the sections are
connected as shown in Figure 123?

(d)

Figure 123: Time taken by the
kettle to boil water in different
configurations.

STUDENT A: First of all we find Figure 123 the total resistance of
the burner for each kind of connection, denoting the resistance
of one section by R. In the initial case (connection in parallel),
the total resistance R 0 = R/ 3. For cases a, b and c (Figure 123 we

WHY DID THE ELECTRIC BULB BURN OUT?

285

obtain

(183)

R

C

2 R 2 2
-= — R

3 R 3

Next, if we denote the voltage applied to the electric table stove
by U, then, using Ohm’s law we can find the total current flow¬
ing through the burner in each case....

TEACHER: (interrupting) You don’t need to find the current. Let
us denote by t 0 , t a , t b and t c the times required to heat the water
in the teakettle to the boiling point in each case. The evolved
heat is equal to the power multiplied by the heating time. In each
case, the same amount of heat is generated. Using formula (182)
to determine the power, we obtain

U 2 t 0 _ U 2 t a _ U 2 t b _ U 2 t c

(184)

Substituting equations (183) into (184) and cancelling the com¬
mon factors ( [U 2 and I/R), we obtain

. U = 2 J± = K

0 3 3 2

from which we readily find the required values:

t a = 9 1 0 = 54 min

ti , = — = 27 mm
b 2

t c = 2t 0 = 12 min

Note that in the given problem it was more convenient to apply
formula (182) to find the power, exactly because the voltage
applied to the electric table stove is a constant value.

But consider the following question. Given: a current source
with an emf $ and internal resistance r; the source is connected
to a certain external resistance R. What is the efficiency of the
source?

286 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: The efficiency of a current source is the ratio of the
useful power, i.e. the power expended on the external resistance,
to the total power, i.e. to the sum of the powers expended on the
internal and external resistances:

r- R R

r, ~ I 2 (R+r) ~ (. R+r)

(185)

TEACHER: Correct. Assume that the internal resistance of the
current source remains unchanged and only the external resis¬
tance varies. How will the efficiency of the current source vary
in this case?

STUDENT B: At R = 0 (in the case of a short circuit), rj = 0.

At R = r, rj = 0.5. As R increases infinitely, the efficiency
approaches unity.

TEACHER: Absolutely correct. And how in this case will the use¬
ful power vary (the power expended on the external resistance)?

STUDENT B: Since the efficiency of the source increases with R,
it follows that the useful power will also increase. In short, the
larger the R, the higher the useful power will be.

TEACHER: You are wrong. An increase in the efficiency of the
current source means that there is an increase in the ratio of the
useful power to the total power of the source. The useful power
may even be reduced. In fact, the useful power is

P

U

g 1 § 2 x

(.R + r) 2 r (x + l) 2

(186)

where x = R/r. If x « 1 then P u ccx. If x » 1 then P u cc—. The

x

useful power P u reaches its maximum value at x = 1 (i.e. R = r),
whenP^ = $ 2 /Ar. A curve of the function y = x/(x + l) 2
is given in Figure 124. It illustrates the variation in the useful
power with an increase in the external resistance.

Consider the following problem. Two hundred identical electric
bulbs with a resistance of 300 Q each are connected in parallel

WHY DID THE ELECTRIC BULB BURN OUT? 287

Figure 124: Graph of the
function y = x/(x + l) 2 which
shows variation in the useful
power with an increase in the
external resistance.

to a current source with an emf of 100 V and internal resistance
of 0.5 Q. Compute the power expended on each bulb and the
relative change in the power expended on each bulb if one of the
two hundred bulbs burns out. Neglect the resistance of the leads
(Figure 125).

STUDENT B: The total current in the external circuit equals

r + -
n

50A

The current passing through each bulb is I t /n = 0.25 A. Next we
can find the power expended on each bulb: P = I"R = 37.5 W.
To determine the relative change In the power per bulb if one
of the two hundred burns out, I shall first find the power P 1 per
bulb for n = 199, and then compute the ratio

/ =

(187)

&

&

•

y R

1

£

1 r

n=Z00

Figure 125: Compute the
power expended on each bulb
and the relative change in the
power expended on each bulb
if one of the two hundred bulbs
burns out.

TEACHER: I do not approve of this method for finding the re¬
quired ratio /. It should be expressed in the general form in
terms of the resistances R and and the number of bulbs n. Thus

288 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Substituting these equations into (187), we obtain

nr +R
nr — r + R

1

nr +R

1

The fraction in the denominator of the last equation is much less
than unity (because there are many bulbs in the circuit and the
resistance of each one is much higher than the internal resistance
of the current source). Therefore, we can apply the approxima¬
tion formula (173):

1 -

r

nr + R

2

- 1

2 r

nr +R

(188)

Alter substituting the numerical values into equation (188) we
find that / = 0.0025.

STUDENT B: But why do you object to computing P 1 first and
then finding / by substituting the numerical values into equation
(187)?

TEACHER: You see that / = 0.0025. This means that if your (nu¬
merical) method was used to obtain this result, we would have to
compute the value of P l with an accuracy to the fourth decimal
place. You cannot even know beforehand to what accuracy you
should compute P l . If in our case you computed Pj to an accu¬
racy of two decimal places, you would come to the conclusion
that power P 1 coincides with power P.

Problems

68. In the electric circuit shown in Figure 126, $ = 100 V, r = 36Q
and the efficiency of the current source equals 50%. Compute the
resistance R and the useful power.

69. A current source is connected to a resistor whose resistance is four
times the internal resistance of the current source. How will the

R

R

R

R

Figure 126: Compute the
resistance R and the useful

power..

WHY DID THE ELECTRIC BULB BURN OUT? 289

efficiency of the source change (in per cent) if an additional-resistor
with a resistance twice the internal resistance is connected in paral¬
lel to the external resistance?

70. Several identical resistances R are connected together in an ar¬
rangement shown in Figure 127. In one case, this arrangement is
connected to the current source at points 1 and 2, and in another,
at points 1 and 3. Compute the internal resistance of the current
source if the ratio of the efficiencies of the source in the first and
second cases equals 16/15. Find the values of these efficiencies.

71. The resistances in a burner of an electric table stove are connected
together in an arrangement shown in Figure 127. This arrangement
is connected to the supply mains at points 1 and 2, and, after a
certain time, 500 grams of water are heated to the boiling point.
Flow much water can be heated to the boiling point in the same
time interval when the arrangement of resistances is connected to
the supply mains at points 1 and 3? The initial temperature of the
water is the same in both cases. Neglect all heat losses.

Figure 127: Compute the
internal resistance of the
current source and hnd the
efficiencies.

72. One and a half litres of water at a temperature of 20 °C is heated for
15 min on an electric table stove burner having two sections with
the same resistance. When the sections are connected in parallel, the
water begins to boil and 100 g of it is converted into steam. What
will happen to the water if the sections are connected in series and
the water is heated for 60 min? The latent heat of vaporization is
539 cal/g. Flow much time will be required to heat this amount of
water to the boiling point if only one section is switched on?

The laws of geometrical optics have been known to mankind for many
centuries. Nevertheless, their elegance and completeness still astonish
us. Find this out for yourself by doing exercises on the construction of
images formed in various optical systems. We shall discuss the laws of
reflection and refraction of light.

§ 31 Do You Know How Light Beams Are Re
fleeted A nd Refracted?

TEACHER: Please state the laws of reflection and refraction of
light.

STUDENT A: The law of reflection is: the angle of incidence is
equal to the angle of reflection. The law of refraction: the ratio
of the sine of the angle of incidence to the sine of the angle of
refraction is equal to the refraction index for the medium.

TEACHER: You statements are quite inaccurate. In the first place
you made no mention of the fact that the incident and reflected
(or refracted) rays lie in the same plane with a normal to the
boundary of reflection (or refraction) erected at the point of
incidence. If this is not specified, we could assume that reflection
takes place as illustrated in Figure 128. Secondly, your statement
of the law of refraction refers to special case of the incidence of a
ray from the air to boundary of a certain medium. Assume that
in the general case the ray falls from a medium with an index
refraction n x on the boundary of a medium with an index of
refraction « 9 . We denote the angle of incidence by a 1 and the
angle of refraction by a 2 . In this case, the law of refraction can be
written as

sinctj « 2
sin a 1 n x

( 189 )

Figure 128: Reflection could
take place in differnt planes.

This leads to your statement provided that for air n i = 1.

294 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Consider the following problem. A coin lies in water at a depth
H. We will look at it form above along a vertical. At what depth
we will see the coin?

STUDENT A: I know that coin will seem to be raised somewhat. I
don’t think I can give a more definite answer.

Figure 129: Where will the
coin appear to be?

TEACHER: Let us draw two rays from the centre of the coin: OA
and OSjS (Figure 129). Ray OA is not refracted (because it is
vertical) and ray OB x B is. Assume that these two diverging rays
enter the eye. The eye will see an image of the coin at the point
of intersection of the diverging rays OA and B^B i.e. at point
Oj. It is evident from the diagram that the required distance h is
related to the depth H by the relation

htana^ = //tana 2

from which we get

, tanas

h = H --

tan a 1

Owing to the smallness of angles aq and a 2 we can apply the
approximation formula

(190)

tan a « sin a « a

(191)

DO YOU KNOW HOW LIGHT BEAMS ARE REFLECTED AND REFRACTED? 295

(in which the angle is expressed in radians and not in degrees).
Using formula (191), we can rewrite equation (190) in the form:

h

H s ' ma 2

sinaj

H

n

(192)

Since for water n = 4/3, we have

4

STUDENT B: What will happen if we look at the coin, not verti¬
cally, but from one side?

TEACHER: In this case, the coin will seem, not only raised, but
moved away (see the dashed lines in Figure 129). Obviously
the computations will be much more complicated in this case.
Consider the following problem. A diver of height h stands
on the bottom of a lake of depth H. Compute the minimum
distance from the point where the diver stands to the points of
the bottom that he can see reflected from the surface of water.

Figure 130: Finding the
minimum distance rom the
point where the diver stands to
the points of the bottom that
he can see reflected from the
surface of water.

STUDENT A: I know how to solve such problems. Let us denote
the required distance by L. The path of the ray from point A
to the diver’s eye is shown in Figure 130. Point A is the closest
point to the diver that he can see reflected form the surface of the
lake. Thus, for instance, a ray from a closed point B is refracted
along the surface and does not return to the diver (see the dashed

296 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

line in Figure 130). Angle a is the critical angle for total internal
reflection. It is found from the formula

1

sinct = — (193)

n

It is evident from the diagram that

L = ht&na + 2(H — h)x.a.na
= (2H — h) tan a

Since tan a =

vT

sin 2 a

then using equation (193) we obtain

L =

2 H-h

\J n 2 — 1

After substituting n = 4/3, we find that

(194)

TEACHER: Absolutely correct. And what kind of picture will the
diver see overhead?

STUDENT A: Directly overhead he will see a luminous circle of a
radius

(H-h)

1 =

V n 2

see Figure 130. Beyond the limits of this circle he will see images
of the objects lying at the bottom of the lake.

STUDENT B: What will happen if the part of the lake bottom
where the diver is standing is not horizontal, but inclined?

TEACHER: In this case the distance L will evidently depend on
the direction in which the diver is looking. You can readily see
that this distance will be minimal when the diver is looking
upward along the inclined surface, and maximal when he looks
in the opposite direction. The result obtained in the preceding
problem will now be applicable only when the diver looks in a
direction along which depth of the lake doesn’t change (parallel
to the shore). A problem with inclined lake bottom will be given
as a homework (see Problem 74).

DO YOU KNOW HOW LIGHT BEAMS ARE REFLECTED AND REFRACTED?

297

STUDENT A: Can we change the direction of a beam by inserting
a system of plane-parallel transparent plates in its path?

TEACHER: What do you think?

STUDENT A: In principle, I think we can. We know that beam,
upon being refracted, travels in a different direction inside a
plate.

STUDENT B: I don’t agree. After emerging form the plate the
beam will still be parallel to its initial direction.

TEACHER: Just prove this, please using a system of several plates
having different indices of refraction.

STUDENT B: I shall take three plates with indices of refraction
n 2 and « 3 . The path of the beam through the system is shown in
Figure 131. For refraction of the beam at each of the boundaries,
we can write

sina 0 sinctj n 2 sina 2 « 3 sinct 3 1

sinoq sina 2 Wj ’ sina 3 n 2 sina 4 « 3 ’

Multiplying together the left-hand sides and right-hand sides of
these equations, respectively, we obtain

sino-p =
sinct 4

Thus a 0 = a 4 , which is what we started to prove.

TEACHER: Absolutely correct. Now, let us discuss the limits of
applicability of the laws of geometrical optics.

STUDENT B: These laws are not applicable for distances of the
order of the wavelength of the light involved and shorter dis¬
tances. At such short distances the wave properties of light begin
to appear.

TEACHER: You are right. This is something that examinees usu¬
ally seem to understand sufficiently well. Can you tell me about
any restrictions on the applicability of the laws of geometrical
optics from the other side - from the side of large distances?

Figure 131: Refracting a beam
through several plates of
different refractive indices.

298 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

STUDENT B: If the distances are longer than the wavelength of
light, then light can be considered within scope of geometrical
optics. At least that is what we were told before. I think there
are no restrictions on the use of geometrical optics for large
distances.

TEACHER: You are mistaken. Just imagine the following picture:
you are sending a beam of light into space, completely excluding
the possibility of its scattering. Assume that in one second you
turn the apparatus sending the beam of light through an angle of
60°. The question is: during this turning motion what will be
velocity of the beam at distances of over 300,000 kilometers from
the apparatus?

STUDENT B: I understand your question. Such points must travel
at velocity greater than that of light. However, according to the
theory of relativity, velocities greater than the velocity of light
are impossible only if they are velocities of material bodies. Here
we are dealing with a beam.

TEACHER: Well, isn’t a light beam material? As you can see,
geometrical optics is inconsistent for excessively great distances.
Here we must take into consideration that a light beam is a
stream of particles of light called as photons. The photons which
were emitted from the apparatus before we turned it “have no
idea” about the subsequent turning motion and continue to
travel in the direction they were emitted. New photons are
emitted in the new direction. Thus we do not observe turning of
the light beam as a whole.

STUDENT B: How can we quantitatively evaluate the limit of
applicability of the laws of geometrical optics from the side of
large distances?

TEACHER: The distances should be such that the time required
for light to cover them must be much less than any characteristic
time in the given problem (for example, much less than the time
required for turning the apparatus emitting the light beam). In
this case, the beam as a whole is not destroyed, and we can safely
use the laws of geometrical optics.

DO YOU KNOW HOW LIGHT BEAMS ARE REFLECTED AND REFRACTED?

299

Problems

73. We are looking vertically from above at an object covered with a
glass plate which is under water. The plate is 5 cm thick; there is a
10 cm layer of water above it. The index of refraction of glass is 1.6.
At what distance from the surface of the water we see the image of
the object?

74. A diver 1.8 m high stands on the bottom of a lake, at a spot which is
5 m deep. The bottom is a plane inclined at an angle of 15°. Com¬
pute the minimum distance along the bottom from the point where
the diver stands to the points on the bottom that he sees reflected
from the surface.

75. We have a glass plate 5 cm thick with an index of refraction equal to
1.5. At what angle of incidence (from the air) will the rays reflected
and refracted by the plate be perpendicular to each other? For this
angle of incidence compute the displacement of the ray due to its
passage through the plate.

76. We have a glass plate of thickness d with an index of refraction
n. The angle of incidence of the ray from the air onto the plate is
equal to the angle of total internal reflection for the glass of which
the plate is made. Compute the displacement of the ray due to its
passage through the plate.

§ 32 How Do You Construct Images Formed By
Mirrors And Lenses?

TEACHER: Quite often we find that examinees are incapable of
constructing images formed by various optical systems, such
as lenses and plane and spherical mirrors. Let us consider some
typical examples. Construct the image of a man formed in the
plane mirror show in Figure 132.

A

Figure 132: Construct the
image of a man formed in the
the plane mirror.

STUDENT A: It seems to me that no image will be formed by the
mirror in this case because the mirror is located too high above
the man.

TEACHER: You are mistaken. There will be an image in the
mirror. Its construction is given in Figure 132 (b). It is quite
evident that to construct the image it is sufficient to prolong the

302 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

line representing the surface of the mirror and to draw an image
symmetrical to the figure of man with respect to this line (surface
of the mirror).

STUDENT A: Yes, I understand, but will the man see his image.

TEACHER: That is another question. As a matter of fact, the man
will not see his image, because the mirror is located too high
above him and is inconveniently inclined. The image of the man
will be visible in the given mirror only to an observer located
within the angle formed by rays A4j and BB V It is appropriate
to recall that the observer’s eye receives a beam of diverging rays
from the object being observed. The eye will see an image of
the object at the point of intersection of these rays or of their
extensions, (see Figure 129 and Figure 132 (b)).

Consider the construction of the image formed by a system of
two plane mirrors arranged perpendicular to each other Fig¬
ure 133 (a).

Figure 133: Construct the
images in the system of two
plane mirror.

(a)

( 6 )

(C)

STUDENT A: We simply represent the reflection of the object
in the two planes of the mirrors. Thus we obtain two images as
shown in Figure 133 (b).

TEACHER: You have lost the third image. Note that the rays
from the object that are within the right angle AOB (Figure 133 (c))
are reflected twice: first from one mirror and then from the
other. The paths of these two rays are illustrated in Figure 133 (c).
The intersection of the extensions of these rays determines the
third image of the object.

HOW DO YOU CONSTRUCT IMAGES FORMED BY MIRRORS AND LENSES?

303

Next, we shall consider a number of examples involving a con¬
verging lens. Construct the image formed by such a lens in the
case illustrated in Figure 134 (a).

STUDENT A: Thats very simple. My construction is shown in
Figure 134 (b).

(d)

(e)

TEACHER: Good. Now assume that one half of the lens is closed
by an opaque screen as shown in Figure 134 (c). What will hap¬
pen to the image?

Figure 134: Examples involving
a converging lens.

STUDENT A: In this case, the image will disappear.

TEACHER: You are mistaken. You forget that the images of
any point of the arrow (for example, its head) is obtained as a
result of the intersection of an infinitely large number of rays
(Figure 134 (d)). We usually restrict ourselves to two rays because
the paths of two rays are sufficient to find the position and size
of the image by construction. In the given case the screen shuts
off part of the rays falling on the lens. The other part of the
rays, however, pass through the lens and form an image of the
object (Figure 134 (e)). Since fewer rays participate in forming
the image, it will not be so bright as before.

STUDENT B: From your explanation it follows that when we

304 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

close part of the lens with an opaque screen, only the brightness
of the image is changed and nothing else. However, anybody
who has anything to do with photography knows that when you
reduce the aperture opening of the camera by irising, i.e. you
reduce the effective area of the lens, another effect is observed
along with the reduction in the brightness of the image: the im¬
age becomes sharper, or more clear-cut. Why does this happen?

TEACHER: This is a very appropriate question. It enables me
to emphasize the following: all our constructions are based on
the assumption that we can neglect defects in the optical system
(a lens in our case). True, the word “defects” is hardly suitable
here since it does not concern any accidental shortcomings of
the lens, but its basic properties. It is known that if two rays,
parallel to and differently spaced from principal optical axis, pass
through a lens, they will, after refraction in the lens, intersect
the principal optical axis strictly speaking, at different points
(Figure 135 (a)). This means that the focal point of the lens (the
point of intersection of all rays parallel to the principal optical
axis), or its focus, will be blurred; a sharply defined image of
the object cannot be formed. The greater the differences in the
distances of the various rays from the principal axis, the more
blurred the image will be. When the aperture opening is reduced
by irising, the lens passes a narrow bundle of rays. This improves
the sharpness to some extent (Figure 135 (b)).

«*)

Figure 135: Image construction
by a lens.

HOW DO YOU CONSTRUCT IMAGES FORMED BY MIRRORS AND LENSES?

305

STUDENT B: Thus, by using the diaphragm we make the image
more sharply defined at the expense of brightness.

TEACHER: Exactly. Remember, however, that in constructing the
image formed by lenses, examinees have every reason to assume
that parallel rays will always intersect at a single point. This
point lies on the principal optical axis if the bundle of parallel
rays is directed along this axis; the point lies on the focal plane
if the bundle of parallel rays is directed at some angle to the
principal optical axis. It is important however, for the examinees
to understand this treatment is only approximate and that a more
accurate approach would require corrections for the defects of
the optical systems.

STUDENT A: What is the focal plane of a lens?

TEACHER: It is a plane through the principal focus of the lens
perpendicular to the principal optical axis. Now, what is the
difference between images formed by a plane mirror and by a
converging lens in the example of Figure 134?

STUDENT A: In the first case (with the mirror) the image is vir¬
tual, and in the second it is real.

TEACHER: Correct. Please explain the differences between the
virtual and real images in more detail.

STUDENT B: A virtual image is formed by intersection, not of the
rays themselves, but of their extensions. No wonder then that a
virtual image can be seen somewhere behind a wall, where the
rays cannot penetrate.

TEACHER: Quite right. Note also that a virtual image can be
observed only form definite positions. In case of a real image
you can place a screen where the image is located and observe
the image from any position. Consider the example illustrated in
Figure 136 (a).

Determine, by construction, the direction of ray AA 1 after
it passes through a converging lens if the path of another ray
in Figure 136 (a)) through this lens is known.

306

QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 136: Construct the
images in the system of two
plane mirror.

STUDENT A: But we don’t know the focal length of the lens.

TEACHER: Well, we do know the path of the other ray before and
after the lens.

STUDENT A: We didn’t study such constructions at school.

STUDENT B: I think that we should first find the focal length of
the lens. For this purpose we can draw a vertical arrow some¬
where to the left of the lens so that its head touches ray BB l .

We shall denote the point of the arrowhead by the letter C (Fig¬
ure 136 (b)). Then we pass a ray from point C through the center
of the lens. This ray will go through the lens without being re¬
fracted and, at a certain point E, will intersect ray Point E
is evidently the image of the point of the arrowhead. It remains
to draw a third ray form the arrowhead C parallel to the prin¬
cipal optical axis of the lens. Upon being refracted, this last ray
will pass through the image of the arrowhead, i.e. through point
E. The point of intersection of this third ray with the principal
axis is the required focus of the lens. This construction is given in

HOW DO YOU CONSTRUCT IMAGES FORMED BY MIRRORS AND LENSES?

307

Figure 136 (b).

The focal length being known, we can now construct the path of
ray AA 1 after it is refracted by the lens. This is done by drawing
another vertical arrow with the point of its head lying on the ray
AA X (Figure 136 (c)). Making use of the determined focal length,
we can construct the image of the second arrow. The required
ray will pass through point and the head of the image of the
arrow. This construction is shown in Figure 136 (c).

TEACHER: Your arguments are quite correct. They are based on
finding image of a certain auxiliary object (the arrow). Note that
this method is convenient when you are asked to determine the
position of the image of a luminous point lying on the principal
axis of the lens. In this case it is convenient to erect an arrow at
the luminous point and construct the image of the arrow. It is
clear that the tail of the image of the arrow is the required image
of the luminous point.

This method, however, is too cumbersome for our example.

I shall demonstrate a simpler construction. To find the focal
length of the lens, we can draw ray EO through the centre of the
lens and parallel to the ray BB l (Figure 136 (d)). Since these two
rays are parallel, they intersect in the focal plane behind the lens
(the cross section of the focal plane is shown in Figure 136 (d) by
a dashed line). Then we draw ray C O through the centre of the
lens and parallel to ray AA 1 . Since these two parallel rays should
also intersect in the focal plane after passing through the lens, we
can determine the direction of the ray AA t through the lens, we
can determine the direction of ray AA l after passing through the
lens. As you can see, the construction is much simpler.

STUDENT B: Yes, your method is appreciable simpler.

TEACHER: Try to apply this method to a similar problem in
which a diverging lens is used instead of converging one (Fig¬
ure 137( a)).

STUDENT B: First I will draw a ray through the centre of lens
parallel to ray BB V In contrast to the preceding problem, the

308

QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Figure 137: Construct the
images with a diverging lens.

extension of the rays and not the rays themselves, will intersect
(we may note for a ray passing through the centre, the extension
will coincide with the ray itself). As a result, the focal plane,
containing the point of intersection, will now be to the left of
the lens instead of the right (see the dashed line in Figure 137 (b)).

TEACHER: (intervening): Note that the image is always virtual in
diverging lenses.

STUDENT B: (continuing): Next I shall pass a ray through the
centre of the lens and parallel to ray AA l . Proceeding from the
condition that the extensions of these rays intersect in the focal
plane, I can draw the required ray.

TEACHER: Good. Now tell me, where is the image of an object a
part of which is in front of the focus of a converging lens, and the
other part behind the focus (the object is of finite width)?

STUDENT B: I shall construct the images of several points of the
object located at various distances from the lens. The points
located beyond the focus will provide a real image (it will be to
the right of the lens), while the points in front of the focus will
yield a virtual image (it will be to the left of the lens). As the

HOW DO YOU CONSTRUCT IMAGES FORMED BY MIRRORS AND LENSES?

309

chosen points approach the focus, the images will move away to
infinity (either to the left or to the right of the lens).

TEACHER: Excellent. Thus, in our case the image of the object
is made up of two pieces (to the left and right of the lens). Each
piece begins at a certain finite distance from the lens and extends
to infinity. As you see, the question “Can an object have a real
and virtual image simultaneously?” should be answered in affir¬
mative.

I see that you understand the procedure for constructing the
images formed by the lens. Therefore, we can go over to a more
complicated item, the construction of an image formed by the
system of two lenses. Consider the following problem:

we have two converging lenses with a common principal optical
axis and different focal lengths. Construct the image of a vertical
arrow formed by such an optical system (Figure 138 (a)). The
focuses of one lens are shown on the diagram by x’s and those of
the other, by blacked in circles *.

W

i

Figure 138: Construct the
images formed in the optical
system.

(b)

(C)

STUDENT B: To construct the image of the arrow formed by two

310 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

lenses, we must first construct the image formed by the first lens.
In doing this we can disregard the second lens. Then we treat
this image as if it were an object and, disregarding the first lens,
construct its image formed by the second lens.

TEACHER: Here you are making a very typical error. I have
heard such an answer many times. It is quite wrong.

Let us consider two rays originating at the point of arrowhead,
and follow out their paths through the given system of lenses
(Figure 138 (b)). The paths of the rays after they pass through
the first lens are easily traced. To find their paths after the second
lens, we shall draw auxiliary rays parallel to our rays and passing
through the centre of the second lens. In this case, we make use
of principle discussed in the preceding problems (parallel rays
passing through a lens should intersect in the focal plane). The
required image of the point of the arrow head will be at the point
of intersection of the two initial rays after they leave the sec¬
ond lens. This construction is shown in detail in Figure 138 (b).
Now, let us see the result we would have obtained if we would
have accepted your proposal. The construction is carried out
in Figure 138 (c). Solid lines show the construction of the im¬
age formed by the first lens; dashed lines show the subsequent
construction of the image formed by the second lens. You can
see that the result would have been entirely different (and quite
incorrect!).

STUDENT B: But I am sure we once constructed an image exactly
as I indicated.

TEACHER: You may have done so. The fact is that in certain cases
your method of construction may turn out to be valid because
it leads to results which coincide with those obtained by my
method. This can be demonstrated on the preceding example by
moving the arrow closer to the first lens i.e. between the focus
and the lens. Figure 139 (a) shows the construction according to
my method, and Figure 139 (b), according to yours. As you see,
in the given case the results coincide.

STUDENT B: But how can I determine beforehand in what cases

HOW DO YOU CONSTRUCT IMAGES FORMED BY MIRRORS AND LENSES? 311

Figure 139: Construct the
images in the system of two
plane mirror.

my method of constructing the image can be used?

TEACHER: It would not be difficult to specify the conditions for
the applicability of your method for two lenses. These condi¬
tions become much more complicated for a greater number of
lenses. There is no need to discuss them at all. Use my method
and you won’t get into any trouble. But I wish to ask one more
question: can a double concave lens be a converging one?

STUDENT B: Under ordinary conditions a double-concave lens is
a diverging one. However, it will become a converging lens if it
is placed in a medium with a higher index of refraction than that
of the lens material. Under the same conditions, a double-convex
lens will be a diverging one.

§ 33 How Well Do You Solve Problems Involving
Mirrors A nd Lenses?

TEACHER: I would like to make some generalizing remarks
which may prove to be extremely useful in solving problems
involving lenses and spherical (concave and convex) mirrors, The
formulas used for such problems can be divided into two groups.
The first group includes formulas interrelating the focal length
F of the lens (or mirror), the distance d from the object to the
lens (or mirror) and the distance f from the image to the lens (or
mirror):

1 11

d + f~F

(195)

in which d, f and F are treated as algebraic quantities whose
signs may differ from one case to another. There are only three
possible cases, which are listed in the following table.

Converging lenses and concave mirrors

Diverging lenses and convex mirrors

d>f

d<f

d>f

d > 0, F > 0 and f > 0

d > 0, F > 0 and f < 0

d > 0, F < 0 and f < 0

Real Image

Virtual Image

Virtual Image

Thus d is always positive; the focal length F is positive for con¬
verging lenses and concave mirrors and negative for diverging
lenses and convex mirrors; and the distance / is positive for real
images and negative for virtual images.

STUDENT A: As I understand, this table enables us to obtain

314 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

three formulas from the general formula (195) which contain the
arithmetical values of the above mentioned quantities:

1 1

Case 1: — + —
a j

1 1

Case 2: — —-
a f

1 1

Case 3: — —-
a f

TEACHER: Yes. Exactly so.

(196)

STUDENT A: Somehow, I have never paid any attention to the
analogy between lenses and the corresponding spherical mirrors.

TEACHER: The second group include formulas which relate the
focal length of the lens (or mirror) to its other characteristics.

For mirrors we have the simple relationship

F = ± | (197)

where R is the radius of curvature of mirror. The plus sign refers
to concave mirrors (the focus is positive) and the minus sign
refers to convex mirrors (the focus is negative). For lenses

)r- (»-l) (2- + 2-) 098)

where n is the index of refraction of the lens material and R t
and i?, are the radii of curvature of lens. If the radius R refers
to a convex side of the lens it is taken with a plus sign; if it refers
to a concave side, with a minus sign. You can readily see that
double-convex, plano-convex and convexo-concave (converging
meniscus) lenses are all converging because, according to formula
(198), they have a positive focus.

STUDENT A: What changes will have to be made in formula (198)
if the lens is placed in a medium with an index of refraction w 0 ?

TEACHER: Instead of formula (198) we will have

(199)

HOW WELL DO YOU SOLVE PROBLEMS INVOLVING MIRRORS AND LENSES? 315

When we pass over from an optically less dense medium ( n 0 < n)
to an optically more dense one (n 0 > n), then, according to
formula (199), the sign of the focus is reversed and therefore
a converging lens becomes a diverging one and, conversely, a
diverging lens becomes a converging one. Let us proceed to the
solution of specific problems.

The convex side of a plano-convex lens with a radius of curvature
R and index of refraction n is silver-plated to obtain a special
type to concave mirror. Find the focal length of the mirror.

STUDENT A: Please allow me to do this problem. We begin
by directing a ray parallel to the principal optical axis of the
lens. After it is reflected from the silver-plated surface, the ray
goes out of the lens and is thereby refracted. If the ray had not
been refracted, it would have intersected the principal axis at a
distance R/2 from the mirror accordance with the formula (197).
As a result of refraction, the ray intersects the principal axis
somewhat closer to mirror. We shall denote the required focal
length by F it is evident from Figure 140 that

316 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

Owing to the smallness of the angles a 1 and a 2 , we can apply the
formula (191). Then

R tan a 2 sin a 2

— =-- m -—- = n

2 F tan sinaq

from which

( 200 )

STUDENT B: I suggest that this problem be solved in a different
way. It is known that if we combine two systems with focal
lengths F l and F 2 , the new system will have a focal length F
which can be determined by the rule for adding the power of
lenses, i.e.

11 1

f~7 1 + y 2

In the given case we have a lens with a focal length

( 201 )

Fi =

R

(»-l)

according to equation (198), where one of the radii is infinite, and
a concave mirror for which

Substituting the expressions for F l and F 2 into formula (201)
obtain

1

F

n — 1

R

2

H-

R

we

( 202 )

from which

F =

R

n + 1

(203)

This shows that STUDENT A did not do the problem right [see
his answer in equation (200)]

TEACHER: (to STUDENT B) No, it is you who is wrong. The
result (200) is correct.

STUDENT B: But is rule (201) incorrect in the given case?
TEACHER: This rule is correct and applicable in the given case.

HOW WELL DO YOU SOLVE PROBLEMS INVOLVING MIRRORS AND LENSES?

STUDENT B: But if rule (201) is correct, then equation (202) must
also be correct.

TEACHER: It is precisely here that you are mistaken. The fact
is that the ray travels through the lens twice (there and back).
Therefore, you must add the powers of the mirror and of two
lenses. Instead of equation (202) you should have written

1 2(n — 1) | 2

F ~ R + R

from which we find that

1 (In —2 + 2)

F ~ R

which leads to

2 n

which coincides with the result obtained in the equation (200).

Consider another problem. A converging lens magnifies the
image of the object fourfold. If the object is moved 5 cm, the
magnification is reduced by half. Find the focal length of the
lens.

STUDENT A: I always get confused in doing such problems. I
think you have to draw the path of the rays in the first position
and then in the second, and compare the paths.

TEACHER: I dare say it will not be necessary at all to draw paths
of the rays in this case. According to the formula (195), we can
write for the given position that

11 1

F~dl + h

Since (fi/d^ = k^) is the magnification in the first case, we obtain

1 _ 1 1 _k x + l

F d x k x d x Ml

317

or

318 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

By analogy we can write for the second position that

d 2 = F

k-, + 1

Thus

d 2 -d l =F k ]- kl (204)

According to the conditions of the problem, d 2 — d 1 = 5 cm, =
4 and k 2 = 2. Substituting these values into equation (204), we
find that F =20 cm.

Problems

77. A lens with a focal length of 30 cm forms a virtual image reduced to
2/3 of the size of the object. What kind of a lens is it (converging or
diverging)? What is the distance to the object? What will be the size
of and distance to the image if the lens is moved 20 cm away from
the object?

78. A luminous point is on the principal optical axis of a concave mir¬
ror with a radius of curvature equal to 50 cm. The point is 15 cm
from the mirror. Where is the image of the point? What will hap¬
pen to the image if the mirror is moved another 15 cm away from
the point?

79. An optical system consists of a diverging and a converging lens
(Figure 141 (a) the x’s indicate the focuses (focal points) of the
lenses). The focal lengths of the lenses equals 40 cm. The object is
at a distance of 80 cm in front of the diverging lens. Construct the
image of the object formed by the given system and compute its
position.

80. An optical system consists of three identical converging lenses with
focal lengths of 30 cm. The lenses are arranged with respect to one
another as shown in Figure 141 (b) (the x’s indicate the focuses
(focal points) of the lenses). The object is at a distance of 60 cm from
the nearest lens. Where is the image of the object formed by the
given system?

(a)

/

\

\

/ /

)

\ '

/

(b)

/

/

\

^ /

\ /

\

\

/ \

/ \

/

Figure 141: Construct the
image formed in the system
and compute its position.

HOW WELL DO YOU SOLVE PROBLEMS INVOLVING MIRRORS AND LENSES?

319

81. The convex side of a plano-convex lens with a radius of curvature of
60 mm is silver-plated to obtain a concave mirror. An object is lo¬
cated at a distance of 25 cm in front of this mirror. Find the distance
from the mirror to the image of the object and the magnification if
the index of refraction of the lens material equals 1.5.

82. The concave side of a plano-concave lens with a radius of curvature
of 50 cm is silver-plated to obtain a convex mirror. An object is lo¬
cated at a distance of 10 cm in front of this mirror. Find the distance
from the mirror to the image of the object and the magnification of
the image if the index of refraction of the lens material equals 1.5.

Answers

322 QUESTIONS AND ANSWERS IN SCHOOL PHYSICS

. 0.43 m/s.

. 27.4 m/s 2 : the direction of the acceleration is
vertically upward.

. 1.28N; 1.28N; 0.62N; 1.56N.

3

. At a distance of — R to the right of the centre
22 5

of the disk.

V

. 0.05 — .

5

. 11.3 cm em; 13.4 mg.

. Lowered by 3 cm; 15.4 mg.

• 59 g.

.(1) 138 J;

(2) 171 J.

. It becomes 1.5 cm longer; 21.5 mg.

. 735 g; it will not be formed; 0.58%.

. 3 x 10“ 8 s; 5 x 10 8 m/s.

. 147 V/m.

( mg+Eq ) (mg + Eq)l s'matzna
cos a ’ 2

(m 1 -m 2 )g+E(q l -q 2 ) ^

(m-i + m 2 )

m l m 2 g + (m 2 q x + m x q^)E
(wj + m 2 )

^5 (mg+Eq) ^ —

1.83 q.

0.16fi.

9.9 V; 1%.

0.196 A; 1.96%.

6 x 10~ 6 C.

SCR X R 2

Aj r + 2 R 2 r +2R l R 1
rdmv^ tana
Sql — dmVq tuna

S'ql — dmVq
( 3(aql

3.75 V; 0.25 A.

60Q;70W.

It is reduced by 28.6%.

J ^

89%; 83%.

100 g of water will be converted into steam;
21 min.

10.8 cm.

I cos a mPsirPa Vcos

/5g/^-

mi

56°; 2.3 cm.

d -(i~-j

n V v » 2 + l/

It is a diverging lens; d = 15 cm; the image will
move 6 cm away from the lens; k = 0.4.

HOW WELL DO YOU SOLVE PROBLEMS INVOLVING MIRRORS AND LENSES? 323

78. f = 37.5cm; the image will become real;
/j = 150 cm.

79. At a distance of 100 cm to the right of the
converging lens.

FINIS

80. It is located at the centre of the middle lens.

81. / = 100cm; k = 4.

82. f = 6.3 cm; k = 0.63.

QUESTIONS
AND ANSWERS
IN SCHOOL PHYSICS

This book was planned as an aid to students preparing for
an entrance examination in physics for admission to an en¬
gineering institute. It has the form of a dialogue between
the author (the Teacher:) and an inquisitive reader (the Stu¬
dent:). This is exceptionally convenient for analysing com¬
mon errors made by students in entrance examinations, for
reviewing different methods of solving the same problems
and for discussing difficult questions of physical theory.

Mir Publishers Moscow